# Blog

## How to Create an Euler’s Method Differential Equations Calculator

Euler’s method is a useful tool for estimating a solution to a differential equation initial value problem at a specific point. In this post, I’m going to show you how to apply Euler’s method both on a piece of paper doing calculations by hand, and in an Excel spreadsheet.

Before we jump into the first example I just want to mention that this uses one of the formulas on my Calculus 2 Study Guide (Integral Calculus Cheat Sheet). It’s available for instant download so you can start using it today. You can learn more about that cheat sheet and buy your copy today by clicking here.

## How to Apply Euler’s Method With Differential Equations

We will go ahead and start with this first example here. Use Euler’s method with a step size of 0.2 to estimate $$y(1)$$, where $$y(x)$$ is the solution of the initial-value problem $$y’=xy-x^2, \ y(0)=1$$. If you’d prefer to see this example in video form you can watch it here.

I like to solve these problems using a table. I think that’s the easiest way to keep everything you’re doing organized. And then we’re going to use the formula on my study guide to fill in the table row by row until we get to our answer. So let’s just go ahead and start with the formula that’s on my study guide first, and then I’ll show you what I mean by setting up the table.

This is just the information that we would need to be given. We know that we have some initial value problem, where we have $$y’ = F(x,y)$$. And then we have the initial condition. So, we know that if we plug in $$x=0$$, into the solution to the initial value problem, we would get out $$y=1$$. Since we know that in this example, we have $$y’ = xy \ – x^2$$, that tells us right there that we’re going to have

$$F(x, y) = xy \ – x^2$$

To apply this Euler’s method formula, what we need to do is set this up in a table. And you can kind of think of this table like an Euler’s method differential equation calculator. I will show you how to use a computer to make this easier. But it is important to know how it works so that you can do it manually too.

### How to Set Up the Euler’s Method Table

We’re going to need a few different columns in our table to keep track of all the calculations. We’re going to start with a column where we keep track of what n we’re on. We also need columns for our $$x_{n-1}$$, and our $$y_{n-1}$$.

Then we also want to calculate what we get when we plug into, our $$F(x_{n-1}, y_{n-1})$$ based on the $$F(x,y)$$ we figured out already. Finally we are going to use all these pieces to figure out our $$y_n$$ based on the formula discussed earlier from my study guide.

It really is just up to personal preference. If you don’t like keeping of all these columns, you don’t really have to. I like to break it down into the smallest possible pieces, and keep track of each individual piece so that you don’t get lost. I do this and recommend this for you because it’s really easy to get lost when you’re trying to keep track of all these different things.

I like to break it down into, at the smallest possible elements of this formula, and keep track of all those, so that when we put it all together, it’s a lot easier to figure out what’s going on. Doing this will give us the following columns to fill in.

First we want to figure out what you need in the n column. The point that we’re given that we start at is, $$x=0$$. So we’re starting at $$y(0)$$. And what we’re trying to estimate is $$y(1)$$. That tells us using the given step size of 0.2, we’re going to start with $$x=0$$ and use Euler’s method to first estimate, what the y value is of this solution when $$x=0.2$$. Then we’re going estimate what the y value is when $$x=0.4$$, then 0.6, then 0.8, and finally when $$x=1$$.

To put it in a more formulaic approach, we would take our ending x value minus the starting x value and divide by the step size.

$$n=\frac{x_n – x_0}{step \ size}=\frac{1-0}{0.2}= 5$$

Well, doing this, is going tell us that we need five steps to get from our starting point to our end. So our n column will have one, two, three, four, and five, because that would represent the five, individual steps that we have to take to get up to $$x=1$$ from $$x=0$$.

### How to Apply Euler’s Method

Now that we have set up our table, we can start applying Euler’s Method to fill the table out. First of all, we need to start with the x and y value that you’re given. We know when $$x=0$$, $$y=1$$. So you’re just going to start with those in the $$n=1$$ row and the $$x_{n-1}$$ and $$y_{n-1}$$ columns.

Then what you can do, is plug these two numbers, $$x=0$$ and $$y=1$$ into the function that we figured out earlier.

$$F(x, y) = xy \ – x^2$$

$$F(0, 1) = (0)(1) \ – (0)^2 = 0$$

And this will go in the $$F(x_{n-1}, y_{n-1})$$ column.

Now with this final column here, $$y_n$$, what we can do is use this formula that we have here, which is on my calculus two study guide. This will use the previous columns along with our given step size of 0.2, which is denoted by h.

$$y_n = y_{n-1}+hF(x_{n-1},y_{n-1})$$

$$y_1 = 1+(0.2)(0) = 1$$

And then we can put this in our final column of this first row.

Now that we figured out this, we can just carry this piece down into the next column. Whatever your previous $$y_n$$ was, is just gonna be your $$y_{n-1}$$ in the next column.

To figure out your next $$x_{n-1}$$, all you have to do is take your previous $$x_{n-1}$$ and just add whatever your step size is. In this case, our step size is 0.2. We’re just going get 0.2 for $$x_{n-1}$$ in the second row of our table. Doing both of these will give us:

### Now Repeat This Process a Few Times

Once you fill out your entire first row, then get the $$x_{n-1}$$ and $$y_{n-1}$$ in your second row, the only thing to do is repeat this process. You will now plug in these two x and y values into the $$F(x_{n-1}, y_{n-1})$$ to get the value in the third column. Then plug those into the formula for $$y_n$$ to get the forth column. Then figure out the $$x_{n-1}$$ and $$y_{n-1}$$ in your third row and continue repeating until your table is full and all 5 rows are filled out. Doing this should leave you with this:

So, after we iterate through this process, all the way up to $$n=5$$, we end up getting in 1.194942 in our $$y_n$$ column. And that should be the answer that they wanted us to find because that should estimate $$y(1)$$.

## How to Apply Euler’s Method in Excel

Turns out, you can also do these Euler’s method problems using Excel or any other spreadsheet software. And it’s a lot easier and faster than doing it by hand. Watch the videos below to see how you can do this yourself to create an Euler’s method calculator to apply Euler’s method in Excel. And don’t forget to get yourself a copy of my integral calculus cheat sheet to help make the rest of your homework and exams easier and smoother!

## Center of Mass Example Problems Using the Center of Mass Equation Integral

Today, we’re going to be going through some center of mass example problems. So, we’re going to sketch the region bounded by the curves and then find the exact coordinates of the centroid of that region. And the curves that we have in this first example are:

$$y=4-x^2, \ and$$ $$y=0$$

If you would prefer to watch a video of this problem, you can do so here:

## Centroid vs Center of Mass

Before we jump into this first example, I do want to just point out one thing. And that is the difference between a centroid vs center of mass. The reason I wanna bring that up is because, you can see this problem is asking us to find the coordinates of the centroid. But I said I was going to show you center of mass example problems. Well, they’re pretty much the same thing in most contexts. In this context specifically, they’re going to be the same.

Really, the only difference when you’re looking at a centroid versus a center of mass is that a center of mass is referring to the center point of some actual physical object that actually has a mass. The context that you’d usually see for that is when you have a thin plate, which is described by some functions or some region, and you want to find the center of mass of that plate. Whereas, a centroid usually comes into play where you are just described some region that exists bound between two curves on an x-y-plane.

Like this case here for example. We were just given these functions and we are looking at the region that is trapped between those functions. Since we’re looking at some region, “centroid” would be the proper terminology there. But if you’re ever given some uniform density object like a thin plate, for example, the centroid and the center of mass are actually going to be the same thing. So you would get the same point whether you were thinking of it in either context. They are going to use the same formulas to figure those out.

So that brings me to the center of mass equation integrals, which are two equations that are on my calculus 2 study guide. If you haven’t checked that out you can click here to learn more about that. You can go download that right away. It’s only a few bucks, it’s pretty affordable. And I highly recommend you grab yourself a copy of that. It’s available right away, so you can go start using that today. Before I show you how to apply the formulas on my study guide though, let’s go ahead and start with graphing the given curves.

## Sketch the Given Curves and the Bounded Region

First we will start with $$y=4-x^2$$. This is just going to be a downward facing parabola with the vertex at the point (0, 4). Then, the line $$y=0$$ is going to be a horizontal line on the x-axis. Therefore, we would get a region like the one below.

So once you’ve sketched your region and kind of given yourself a visualization of what we’re trying to do here, the best place to go from there is to just go straight into the center of mass equation integrals.

## Center of Mass Equation Calculus

There’s a separate equation for the x-coordinate of the centroid and for the y-coordinate of the center of mass. Those equations, from my calculus 2 study guide are:

$$\bar{x} = \frac{1}{A} \int_a^b x f(x) \ dx$$ $$\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} \Big[ f(x) \Big]^2 \ dx$$

Where the centroid of the region will be at the point $$( \bar{x}, \ \bar{y} )$$.

Before I show you how to use these, I do just wanna point out the different pieces of these equations. First of all, we have A in both of these equations. A is just the area of the region whose centroid, or center of mass, we’re trying to find. We would first need to figure out the area between these two curves. And that’s what A would be. I’m not going to show you how to do that here, but you can see more about that by clicking here.

Then we have the bounds of our integrals, the a and b. Those bounds of the integrals are just going to be the x values that are the left and right edge of our region. So in this case, a will be -2. Meanwhile, b will be 2. And that’ll be true for both of those integrals.

Finally, we have f(x). That is just going to be the function that creates this region. These functions assume that the lower bound of the region will be formed by the line $$y=0$$. Based on that, f(x) is just going to be the top function, $$y=4-x^2$$.

You can simply use those pieces in the formulas above, and I’ll show you how to do that shortly. However, before we do that I do want to point out one thing.

## Finding the Center of Mass of a Symmetrical Region

This is an interesting example, because you can see this region whose centroid we are looking for, is actually symmetrical in the x direction. This region is symmetrical to the right and to the left of the y-axis. Whenever you have symmetry in your region that actually saves you some work. Since we have symmetry in the x direction, that tells us that the x coordinate of our centroid must be on that line of symmetry, which is the line $$x=0$$. As a result, we already know that we’ll have $$\bar{x} = 0$$.

Since we already know $$\bar{x}=0$$, all we need to do is use the above formula to calculate $$\bar{y}$$.

## How to Apply the Center of Mass Formula

If you use that method shown here, you would figure out that the area of this region is $$\frac{32}{3}$$. And like I said earlier, the bounds of the integral are the left edge and the right edge of this area. So that gives us $$a=-2$$ and $$b=2$$. I also mentioned above that we will know that $$f(x)=4-x^2$$. Plugging this into the $$\bar{y}$$ equation tells us:

$$\bar{y} = \frac{1}{32/3} \int_{-2}^{2} \frac{1}{2} \Big[ 4-x^2 \Big]^2 \ dx$$

I do want to point out something important about the above equation. When you are applying these equations you put the f(x) all in brackets or parentheses, because we need make sure to square this whole function. From here we can simplify things a bit. When you have a fraction in the denominator of another fraction like we do here, you want to keep in mind that dividing is the same as multiplying by its reciprocal. Therefore, $$\frac{1}{32/3}$$ can be rewritten as $$\frac{3}{32}$$.

Then we can expand and simplify the rest of the integral.

$$\bar{y} = \frac{3}{32} \cdot \frac{1}{2} \int_{-2}^{2} (4-x^2)(4-x^2) \ dx$$

$$\bar{y} = \frac{3}{64} \int_{-2}^{2} 16-8x^2+x^4 \ dx$$

$$\bar{y} = \frac{3}{64} \Bigg[ 16x \ – \frac{8}{3}x^3+ \frac{x^5}{5} \Bigg]_{-2}^2$$

$$\bar{y} = \frac{3}{64} \Bigg[ \Bigg( 16(2) \ – \frac{8}{3}(2)^3+ \frac{(2)^5}{5} \Bigg) – \Bigg( 16(-2) \ – \frac{8}{3}(-2)^3+ \frac{(-2)^5}{5} \Bigg) \Bigg]$$

$$\bar{y} = \frac{3}{64} \cdot \frac{512}{15}$$

$$\bar{y} = \frac{8}{5}$$

That tells us that the y-coordinate of our centroid of this region is $$\bar{y}=\frac{8}{5}$$. And we already figured out that the x-coordinate of our centroid was $$\bar{x}=0$$. Therefore, the centroid of this region between these two given functions is going to be $$(0, \ \frac{8}{5})$$.

Again, the center of mass equations are on my calculus two study guide. You can click here to check that out and get your copy today!

## How to find the work required to pump the water out of the tank

In this post, I am going to be showing you how to find the work required to pump the water out of the spout.  Work is a common topic in calculus 2, and there are a lot of different applications for it.  But finding the work required to pump the water out of a tank is one of the most common applications.

It is also important to keep in mind that the process is slightly different depending on whether your tank is measured in feet or inches versus meters.  So, I will show you how both work out.  But keep in mind that the process is extremely similar except for one main difference, which I will talk more about more soon, and the units of course.

## Find the work required to pump the water out of the spout – Meters

If you are given some sort of a tank like in this case, a spherical tank that we can see has a radius of 3 meters.  And then there is the spout sticking out the top of it, which is 1 meter long.

This is going to use one of the formulas mentioned on my calculus 2 study guide.  You can click here to check that out, but you can go download that right away.  It is available for instant download; it will be a PDF that you can print or save.  It is only a couple bucks, so it is very affordable, but this is one of the formulas on there.

### The General Strategy

When it comes to figuring out the work required to pump water out of a tank, really what that comes down to is setting up an integral which represents the work required to do this.  But when you are setting that integral up, you have to go about it step-by-step and work your way through the different steps.  We are not going to go straight to the formula for work quite yet.  What we want to do first is imagine, as water is being pumped out of this tank, what is going to be happening is, you can imagine a bunch of little disks.  You can see one such disk in the image below.

Each little disk is basically an infinitely thin cylinder, which represents all the water that sits in that layer of our tank.  So, you can kind of imagine, instead of thinking of this as a sphere, think of it as a bunch of really thin cylinders stacked on top of each other that make the shape of a sphere.  And the reason why we want to think of it like this is each disk of water in this tank, is going to require a certain amount of work since all the water in that layer must be pumped up the same distance.  All the water on this layer is going to be the same distance from the top of the spout where we are trying to get that water to.  So basically, what we are going to be doing, is figuring out an equation which represents the amount of work required to pump a specific disk out of this tank.  And then we are going to sum up all those works that it takes to get each disk, and that be the work that it takes to get all the disks of water out of the tank.

### Draw a Sketch and Label Your Variables

To do this, what we want to do, is come up with a new variable, which represents the distance that any given layer of water has to be pumped out.  That distance is going to be the distance from whatever layer of water we are looking at up to the top of the spout, which is the distance labeled $$x_i^*$$.

And you can imagine, these different disks as we go throughout this whole thing, are all going to have a different distance that they need to be lifted.  The disk down at the bottom of the tank is going to have to be pumped further than the disk up at the top of the tank.  Once we have this variable that represents the distance that the ith layer has to be pumped up, what you want to do is figure out an equation for the volume of that layer.

### Volume of the ith Layer

We want to come up with an equation for the ith layer of the liquid.  This is going to be based on $$x_i^*$$ because, you can see as you go throughout this sphere, the radius of the individual disk changes.  The distance from the center of the disk to the edge of the disk is much shorter at the top and the bottom of this sphere, and much longer in the middle of the sphere.  In the middle, we know it is 3 meters, but closer to the top or bottom of the tank it is going to be shorter. The volume of each disk is going to be different depending on how far down into the tank you are.

Because the volume of each individual disk is just going to come from the volume of a cylinder equation, we can start with that as a template and adjust from there.

$$V=\pi r^2 h$$

The radius is going to depend on how far down into the tank we are.  So, what we need to do is come up with an equation which depends on $$x_i^*$$, and represents the radius of that specific disk sitting at the depth of $$x_i^*$$.  To do that we can graph this circular shape and spout on an x-y-axis.  We will say that the top of the spout is at the origin and will graph the depth from the top of the spout ($$x_i^*$$) on the x-axis, and the radius of the tank at that depth (r) on the y-axis.

What we want to do is figure out an equation for that circle.  Generally, the equation for a circle is going to be $$(y-y_0)^2+(x-x_0)^2=r^2$$.  Since the center of this circle is at (4, 0) and the radius of the tank is 3 m, we can plug in these values to see that the equation of this circle is

$$(y-0)^2+(x-4)^2=3^2$$

Then we can simplify this equation and solve for y, leaving us with

$$y=\pm \sqrt{9-(x-4)^2}$$

So if we want a formula for the radius of our disk we can just take the positive square root piece, since that represents the top half of the circle.  Therefore, the radius is going to be $$y=\sqrt{9-(x-4)^2}$$.  But we’ll want to change the x to $$x_i^*$$. So the volume of our ith disk is just going to be

$$V_i = \pi \bigg( \sqrt{9-(x_i^*-4)^2} \bigg) ^2 \ h$$

Now we need to figure out the height. Well, the height of each of these disks will simply be $$\Delta x$$.  All that means is the distance that you go from one disk, or one layer, of your water to the next.  If we make this change, that tells us

$$V_i = \pi \bigg( \sqrt{9-(x_i^*-4)^2} \bigg) ^2 \ \Delta x$$

And we can simplify this a bit.

$$V_i = \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

Once we have the volume of the ith disk, what we needed to do is figure out the mass of that disk.

### Mass of the ith Layer

Once we have found an equation for the volume of the ith layer of water in the tank, finding the mass of the ith layer is fairly straight forward. The reason for this is that the mass of the layer will just be the volume of the layer times the mass of water. Since we know the mass of water to be $$1000 \frac{kg}{m^3}$$, this tells us that

$$m_i = 1000 * V_i$$

$$m_i = 1000 \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

### Force Acting on the ith Layer

Again, finding the force acting on the ith layer is pretty simple once we know the mass of the ith layer. This is because the force is mass times gravity. Or in other words

$$F_i = g * m_i$$

Since we know the gravitational constant for the strength of gravity on Earth is $$9.8 \frac{m}{s^2}$$, we just have to multiply the mass of the ith layer by 9.8 to get the force acting on the ith layer.

$$F_i = 9800 \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

### Work Required to Pump the ith Layer Out of the Spout

Now that we know the force acting on the ith layer, we can use this to find the work required to pump the ith layer of water up and out of the spout. This is where the formula on my calculus 2 study guide comes in. That formula says that work is force times distance.

Since we already know the force of the ith layer, the only thing left to figure out is the distance that the ith layer needs to be pumped. Well, if we look back up to the drawing where we named our variables, you will see that $$x_i^*$$ represents the distance between the ith layer of water and the top of the spout. So $$x_i^*$$ is the distance that the ith layer needs to be pumped. So if we know that

$$Work \ = \ Force \ * \ Distance$$

And that $$x_i^*$$ is the distance that the ith layer needs to be pumped, then

$$W_i = F_i \ * \ x_i^*$$

$$W_i = 9800 x_i^* \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

### Total Work

At this point we only know the work required to lift the ith layer, not the work that is required to lift all the water out.  What we need to do now is use this to set up an integral which would give us the total amount of work to lift all the i layers, which is all the water.  To do that we can set up an integral. And we are going to integrate this work that it takes to lift the ith layer.  Basically, what the integral is doing is summing up all the amounts of work that it takes to lift each individual layer. It is adding all those up together to give us one total amount of work that it takes to pump all the water out of the spout.

When you put it in terms of an integral, this $$\Delta x$$ changes to a dx.  We would also need to change the $$x_i^*$$ to x.  This change of variable just changes us from being in the context of a sum to the context of an integral, which is just a special type of an infinite sum itself.

$$\int 9800 x \pi \Big( 9-(x-4)^2 \Big) \ dx$$

### But what about the bounds?

Now what we need to figure out is the bounds of our integral.  The bounds are going to represent the range of these x values, that we want to integrate over.  To do this we want to think about where the x came from. Again, this was the distance that we must pump our water.  What we need to figure out is: what is the range that we have to pump each layer of water?  What are the total range of distances?

Well, we can see looking back up to our drawing, that the top layer of water is already at the bottom of the spout.  It just has to be pumped up 1 meter to get to the top of the spout.  So, the shortest distance we are going to have to pump is 1 meter. That tells us that the lower bound of our integral is going to be 1.

The very, very bottom bit of water down at the bottom of our tank is going to have to be pumped all the way up through the tank and then all the way up through the spout.  If the radius of our sphere is 3 meters, it is going to have to go 3 meters to get up to the center of our sphere, another 3 meters to get up to the top of the sphere, and then another 1 meter to get to the top of the spout.  So, that is going to be $$3 + 3 + 1 = 7$$ total meters.  That tells us that the upper bound is going to be 7.

$$\int_1^7 9800 x \pi \Big( 9-(x-4)^2 \Big) \ dx$$

So now we can integrate this and that would give us the total work that it would take to pump all the water out of the spout.

$$9800 \pi \int_1^7 x \Big( 9-(x-4)^2 \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( 9-(x-4)(x-4) \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( 9-(x^2-8x+16) \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( -x^2+8x-7) \Big) \ dx$$

$$9800 \pi \int_1^7 -x^3+8x^2-7x \ dx$$

This can be done using the power rule, then evaluating over our bounds of 1 to 7, we get $$1,411,200 \pi$$. So, we know that it would take $$1,411,200 \pi$$ Joules worth of work to pump all of the water out of the spherical tank.

So, like I said, this is one of the formulas on the Jake’s Math Lessons calculus 2 study guide, click here to go check that out and download your copy today.

## Is Cady Right, or Does the Limit Exist?

I recently watched the movie Mean Girls. At the end of the movie, there is a scene where the main character is shown a limit and asked to evaluate it. After quickly working out the problem in a tight competition for the academic decathlon, Cady looks up from her paper and shouts, “THE LIMIT DOES NOT EXIST!”

And of course, being the math oriented person that I am, I just was left wondering whether the limit actually exists after Mean Girls concluded. So I figured I’d share my findings.

First of all, here’s the limit shown at the end of Mean Girls: $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)}$$

## Where do we start?

Typically when I’m evaluating limits, I like to assume that the function is continuous at the point we’re evaluating the limit and just plug it in. This usually doesn’t work, but it will frequently give us a hint as to which method we can use to evaluate it. This basically just takes advantage of the basic limit properties of limits if it works out. But we need to make sure we’re not causing any problems when we do this.

So let’s just start with plugging in zero for x in the function who’s limit we’re looking for.

$$f(x) = \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)}$$ $$f(0)=\frac{ln(1-0) \ – sin(0)}{1-cos^2(0)}$$ $$f(0)=\frac{ln(1-0) \ – sin(0)}{1- \big( cos(0) \big)^2}$$ $$f(0)=\frac{0 \ – 0}{1- 1} = \frac{0}{0}$$

Turns out, this doesn’t really work this time. Because we end up with the indeterminate form of $$\frac{0}{0}$$. Since it’s never fine to divide by zero, this isn’t allowed.

So, we’ll need to think of another way to evaluate this limit. But like I said before, the fact that we got this indeterminate form gives us a hint of how we can evaluate this limit properly. This indeterminate form actually is one of the conditions that tells us we can evaluate this limit using L’Hospital’s Rule.

## L’Hospital’s Rule

Which means we can create a new limit. We will still have $$x \to 0$$ in our new limit. But we’ll take the limit of a different function. And hopefully this one will be easier to evaluate.

This new limit will actually just be made up of a fraction where the top of the fraction is just the derivative of the original numerator. And the bottom of this new faction is just the derivative of the denominator of the original fraction. So we know that $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)} \ = \ \lim_{x \to 0} \ \frac{\frac{d}{dx} \big[ ln(1-x) \ – sin(x) \big] }{\frac{d}{dx} \big[ 1-cos^2(x) \big] }$$

Remember, we are not finding the derivative of the fraction as a whole, so we should not use the quotient rule. We will need to apply the chain rule to find the derivative of a few of these terms. Doing so tells us that $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)} \ = \ \lim_{x \to 0} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$

## Now can we evaluate this limit?

The point of using L’Hospital’s Rule is that we should end up with a limit that’s easier to evaluate. So, let’s try the same thing we did earlier and see what happens. Since we have a limit as $$x \to 0$$, let’s just plug zero in for x and see what we get. $$\lim_{x \to 0} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ $$\frac{ – \ \frac{1}{1-(0)} \ – \ cos(0) }{ 2cos(0) \cdot sin(0)}$$ $$\frac{ – \ 1 \ – \ 1 }{ 2(1)(0)} = \frac{-2}{0}$$

And you can see we’ve divided by zero. So, this isn’t going to work. You can’t divide by zero, it breaks the rules of math. Therefore, we can’t just evaluate this limit by plugging in zero for x.

However, we are getting closer. The reason I say this is that we don’t have the indeterminate form of $$\frac{0}{0}$$ anymore. Since we have some other number divided by zero, that tells us that this limit will likely be either $$\infty$$ or $$– \infty$$.

The reason for this is that the numerator of the fraction is going toward the number -2. Whether you approach from the left or the right of x = 0, the top of our fraction is going to be a number close to -2.

Meanwhile, as x gets really, really close to 0, the denominator is going to also get really close to 0. But we don’t care about what the denominator is when x = 0. Since we’re dealing with a limit we care about the denominator when x is really close to 0. But depending on which side of zero we’re on, the denominator could be positive or negative.

If we have a fraction where the numerator is some non-zero number, and the denominator is getting infinitely close to zero, the fraction as a whole will become infinitely large. We just need to figure out if it’s positive or negative. Well, we can figure this out by splitting the limit up into two one-sided limits and see what those tell us about the two-sided limit.

## Left-Sided Limit

Let’s start with the limit where x is approaching zero from the left. $$\lim_{x \to 0^-} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ We already know that the numerator of this fraction is going to approach -2 as x approaches 0 from either side. So, let’s just look at the denominator of this fraction.

What happens to $$2cos(x)sin(x)$$ as $$x \to 0$$? Well, we already said it will approach 0, but let’s just consider what’s going on as x approaches 0 from the left. If we’re approaching 0 from the left, that means we want to consider x values that are super close to 0, but will be negative. For negative x values approaching 0, cos(x) will approach 1, and will be slightly smaller than 1. And sin(x) will approach 0, and will be negative numbers very close to 0.

This is because sin(x) is negative for x values very close to 0. We can see this looking at a graph of $$f(x)=sin(x)$$.

Based on this, as x approaches 0, 2cos(x)sin(x) will become the product of 2, 1, and a negative number very close to 0. The product of two positive numbers and a negative number will be a negative number. So we know that this denominator will be getting closer and closer to 0, but will be a negative number in this left sided limit.

So we can think of the fraction as a whole, as a positive number very close to 2, divided by a negative number very close to 0. As x goes to 0 from the left, this fraction will get infinitely large and will be negative because it’s a positive divided by a negative. In other words, as $$x \to 0^-$$, $$\frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) } \to \frac{2}{-0}$$. Or $$\lim_{x \to 0^-} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }=-\infty$$

## Right-Sided Limit

The right-sided limit will work out very similarly, but with one main difference. $$\lim_{x \to 0^+} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ Again, the numerator of this fraction is going to approach -2 as x approaches 0 from the right. But let’s look at the denominator.

By the same process as the left-sided limit above, we can see that the denominator of this fraction is going to approach 0. But what we need to figure out is whether it is going to be approaching 0 from the negative side or the positive side. And the only term of our denominator that is going to be different approaching from the right side of 0, is sin(x).

Looking back up at the graph of $$f(x)=sin(x)$$ you can see that the function is positive (above the x axis) to the right of $$x=0$$. Therefore, the denominator of this fraction is going to be the product of three positive numbers, resulting in a positive number. And the fraction as a whole will be a positive number over a positive number. In other words, as $$x \to 0^+$$, $$\frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) } \to \frac{2}{+0}$$. Or $$\lim_{x \to 0^+} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }=\infty$$

## What does this tell us about the two-sided limit?

Remember, for a limit to exist both one-sided limits need to exist AND they both need to be equal to each other. We just figured out that our left-sided limit is $$-\infty$$ and the right-sided limit is $$\infty$$. Since one is positive and one is negative, they clearly aren’t equal. Each one-sided limit does not give the same result. Therefore, the limit we are trying to find indeed does not exist.

So, it turns out Cady was right in the end of Mean Girls. The limit does not exist.

## The Calculus Lifesaver by Adrian Banner Review

Taking a very causal and easy to understand approach, Banner has created a complete guide for any introductory calculus student.  This book really does have all the tools you need to excel at calculus.  It is titled appropriately I would say.

More important than that though, is that this book is put into simple terms making it easy to understand.  It is common to see calculus textbooks that use technical mathematical language to explain complicated concepts.  The Calculus Lifesaver: All the Tools You Need to Excel at Calculus (Princeton Lifesaver Study Guides) by Adrian Banner is not one of those books.  Adrian Banner uses casual language that is easy to understand.  I think this is important because anyone that is trying to learn calculus topics for the first time would not be an expert on the technical jargon in order to make sense of the concepts as they are described in some other textbooks.

## How much does The Calculus Lifesaver cost?

The Calculus Lifesaver is priced very reasonable.  Below is the pricing available at the time of this writing, but you can confirm the updated cost on Amazon by clicking here.

You can purchase both the e-book and paperback version of this book on Amazon for under $20. The price for ebook and paperback tend to stay fairly consistent, whereas the hardback price really ranges. If you prefer hardback instead, it will cost you a little bit more, coming in between$20 – \$80.  If you want to get access to this book for even less, Amazon does offer it used and to rent if you only need it for the term you are taking a calculus class.

## Who is this book not helpful for?

Although this casual approach can be great for someone learning the basic concepts on which calculus is built, this could be a negative for a math major who will need to go more in depth or someone trying to develop a deeper understanding of the theory or thought behind the math.  If you are interested in proofs and understanding how calculus was derived, this probably is not the book for you.

Another downside of this book is that it does not contain collections of problems at the end of each chapter.  It is not packed full of practice problems for you to practice the concepts discussed in the chapters.  There are certainly problems that are solved by Banner throughout the book, but there is not a ton of problems that you would be able to do on your own.  It is more of an explanation of the concepts you need to know as well as an explanation of how to apply them.  Then you would be left on your own to find problems that you can get some practice with.  It does fantastic job at breaking down all the tools you need to know to excel at calculus and how to apply them, then sends you on your way to go get the repletion you need elsewhere.

Fortunately, there are other options out there to get the list of practice problems.  One option could be to look in the required textbook for your course if you are enrolled in one.  If you are not taking a calculus class though, there are other books that you can get that are just books of practice problems.  If you pair one of those with The Calculus Lifesaver, you should be set to make sure you know the concepts you need and get the practice to really get it down solid.

## Who is this book perfect for?

Like I mentioned before, Banner takes a very casual approach in this text.  He does not focus on the technical mathematical terminology that most textbooks focus on much more.  This fact is certainly a positive for someone learning calculus for the first time.  If math is not your major and you are taking a calculus class for general requirements this could be the perfect book for you.  It’s the perfect bridge to get you from being new to calculus to understanding the complex explanations in the textbook you may have already had to buy.

For example, at one point in this book, Banner writes, “perhaps you’d like to say that f(2)=1, but that would be a load of bull since 2 isn’t even in the domain of f.”  This is just one example that illustrates the casual approach Banner takes in The Calculus Lifesaver.  And there are many more.

And I am not trying to say that the technical terminology in math is not important.  I think it certainly is.  But it is usually helpful to understand the concepts before trying to build on that with a more in depth understanding of the language and terminology.  When you are learning about a brand-new topic in math, it can be easy to get distracted with the terms.  And this can prevent you from putting more focus on the general concepts that you are trying to understand.  This may make you feel like you don’t understand the concepts you are trying to learn, when in reality you would understand the concepts it they were explained in terms you already understand.

This pickle is completely avoided thanks to the casual approach in The Calculus Lifesaver.  Banner ensures that the language is simple and straight forward so you can easily understand the thinking behind the actual math.  It is done brilliantly.

## What topics are covered in The Calculus Lifesaver by Adrian Banner?

The Calculus Lifesaver covers topics that you need to know in calculus 1, 2, and 3.  You know you will really be able to get some longevity out of this book if you need to take multiple calculus classes.  It’s always great when you can buy one book that can be used to help you through multiple courses.  This is not a comprehensive list, but some of the main important topics covered by Banner in this book are:

• Functions and review of prerequisite material
• Limits and techniques for evaluating them
• Continuity
• Derivatives and different methods to find them
• Optimization and linear approximation
• Implicit differentiation and related rates problems
• Definite and indefinite integrals and how to find them
• The Fundamental Theorem of Calculus (FTC)
• Improper integrals
• Basic concepts of sequences and series
• Taylor Polynomials, Taylor Series, and Power Series
• Polar coordinates and how they relate to topics of calculus
• Volumes, arc length, and surface area
• Differential equations

## Final Recommendations

I strongly recommend this book for anyone that is a non-math major taking calculus classes because they need to for their major.  I would also recommend this book for any high school students taking AP Calculus AB or BC or taking IB Math SL or HL.  It may be more applicable for an IB Math HL student than SL, but it would certainly be helpful for both.  I think this book would also be immensely helpful for someone who is trying to learn calculus without having enrolled in a high school or college calculus class.  The casual approach is perfect for anyone that is trying to get a grip on the basic concepts taught in calculus and how to apply them.  Click here to go get yourself a copy of The Calculus Lifesaver.

However, if you are a math major or are interested in learning more about the ideas behind calculus or proving the concepts that make up calculus, this may not be the book for you.  This is certainly not a proof-based resource and does not contain a ton of practice problems.  So if you are looking to really dive deep into calculus, you may want to consider Calculus by Michael Spivak, which would be my personal recommendation for a proof-based approach to calculus.

Some links contained in this post are affiliate links, meaning I would get a small commission for your purchase at no additional cost to you.

## How to Find the Integral of e^x+x^e

$$\int e^x + x^e \ dx$$

Finding the integral of $$\mathbf{f(x)=e^x+x^e}$$ can be tricky at first glance. I’m sure you’re probably familiar with how to take the integral of the $$\mathbf{e^x}$$ part of it. In general, you’ll just want to remember that $$\int e^x \ dx = e^x$$

However, the $$\mathbf{x^e}$$ piece looks a little weird. At first glance it may even look completely foreign. What have you seen that looks like this piece of the function?

## Let’s think about what’s going on here

If we look back at our original integral, you’ll want to notice the dx at the end of it. Remember we had $$\int e^x + x^e \ dx$$

The dx is important because it indicates what letter is our variable that we will be integrating with respect to. Since it’s dx, that means we will be integrating with respect to x. Therefore, e will need to be treated as a constant.

In fact, e is a known constant with a specific value. It’s kind of like $$\mathbf{\pi}$$. We know that $$\mathbf{e \approx 2.71}$$. So when we are trying to integrate the $$\mathbf{x^e}$$ part of this function, what we are integrating just comes down to “a variable raised up to a constant power.”

When you are trying to integrate x raised up to some constant power, you would want to use the power rule. The power rule for integrating tells us that if n is some constant other than n = -1, then $$\int{x^n} \ dx \ = \ \frac{x^{n+1}}{n+1} \ = \ \frac{1}{n+1} \cdot x^{n+1}$$

So since e is a constant, we can basically just replace the n in the above formula with e to find the integral of $$\mathbf{x^e}$$. Using this, we can see that $$\int{x^e} \ dx \ = \ \frac{x^{e+1}}{e+1} \ = \ \frac{1}{e+1} \cdot x^{e+1}$$

## Putting these integrals together

Now that we have figured out how to integrate the $$\mathbf{x^e}$$ part of our function, let’s go back to the original integral. As we do this, let’s also think about one of the basic integral properties. Specifically, the one that tells us what to do about integrating a function that is a sum of two simpler functions. $$\int f(x)+g(x) \ dx \ = \int f(x) \ dx + \int g(x) \ dx$$

So as a result of this, we can break down our problem into two simpler integrals that we already know. $$\int e^x + x^e \ dx \ = \int e^x \ dx + \int x^e \ dx$$

And since we already found both of these integrals earlier, we know $$\int e^x + x^e \ dx \ = \ e^x + \frac{x^{e+1}}{e+1} + C$$

## Linear Approximation (Linearization) and Differentials

Linear approximation, sometimes called linearization, is one of the more useful applications of tangent line equations. We can use linear approximations to estimate the value of more complex functions. Coming up with a linear function that closely approximates another function at a certain point gives you something that is a lot easier to work with than the original function.

## How to find a linear approximation at a point

To find the linear approximation of a function at a point, you can simply use the formula for the linearization of a function. Let’s say you are given a function $$f(x)=x^3+2x$$ and you want to find its liner approximation at the point $$\mathbf{a=-2}$$. You can do this by applying the formula

$$L(x)=f(a)+f'(a)(x-a).$$

The first step is to find the derivative of f(x). In this case we can find f'(x) by using the power rule.

$$f'(x)=3x^2+2$$

Now that you know f(x), f'(x), and the value of a, you can plug all of this into the linearization formula above to find the linear approximation of f(x) near x=a.

$$L(x)=f(a)+f'(a)(x-a)$$ $$L(x)=\big( (-2)^3+2(-2) \big)+ \big( 3(-2)^2+2 \big) \big( x-(-2) \big)$$ $$L(x)=\big( -12 \big)+ \big( 14 \big) \big( x+2 \big)$$ $$L(x)=-12 + 14x+28$$ $$L(x)=14x+16$$

## Useful Applications of Linear Approximation

At this point you’re probably thinking, “this is just finding tangent line equations, who cares?”

But this does have some useful applications. One such application is using linear approximation to estimate the value of numbers that would be difficult or impossible to find without a calculator.

Let me show you what I mean by that.

### Example

Find the linearization of $$\mathbf{f(x)=\sqrt{x+1}}$$ at a=0 and use it to approximate $$\mathbf{f(x)=\sqrt{0.9}}$$.

Finding $$\mathbf{f(x)=\sqrt{0.9}}$$ would be difficult without a calculator, right? But with linear approximation we can get pretty close by applying the process above and taking it a step further pretty easily.

We already have been given f(x) and the a value we need, so you just need to find f'(x) to apply our linearization formula.

$$f(x)=\sqrt{x+1}$$ $$f(x)=(x+1)^{\frac{1}{2}}$$

Writing f(x) in this form, we can simply apply the power rule and chain rule to find f'(x).

$$f'(x)=\frac{1}{2}(x+1)^{-\frac{1}{2}}$$ $$f'(x)=\frac{1}{2\sqrt{x+1}}$$

Now that we know f'(x), let’s first use that and f(x) to calculate f(a) and f'(a) before applying all of this to the linearization formula. Remember we were given a=0.

$$f(x) = \sqrt{x+1}$$ $$f(0)=\sqrt{0+1}$$ $$f(0)=\sqrt{1}$$ $$f(0)=1$$

And now you can find f'(a), or f'(0).

$$f'(x)=\frac{1}{2\sqrt{x+1}}$$ $$f'(0)=\frac{1}{2\sqrt{0+1}}$$ $$f'(0)=\frac{1}{2\sqrt{1}}$$ $$f'(0)=\frac{1}{2}$$

Now we can plug all of this into our linearization formula.

$$L(x)=f(0)+f'(0)(x-0)$$ $$L(x)=1+\frac{1}{2}x$$

#### How do you use this to estimate a number?

So we know that the function $$L(x)=1+\frac{1}{2}x$$ is a good estimate of $$f(x)=\sqrt{x+1}$$ when we plug in numbers close to a=0 for x. And if we plug x=0 into f(x), that would give us $$f(0)=\sqrt{0+1}=\sqrt{1}$$ which is close to the number we are trying to estimate, $$f(x)=\sqrt{0.9}$$.

In reality, we could find the exact value of $$\sqrt{0.9}$$ by evaluating $$f(-0.1)=\sqrt{-0.1+1}=\sqrt{0.9}$$. But this is difficult, so instead we can use our linear approximation because we know it’s close to f(x) for x values near x=0. So instead of finding f(-0.1) we can find L(-0.1) and that will give us a good estimate.

$$L(-0.1)=1+\frac{1}{2}(-0.1)$$ $$L(-0.1)=1+(0.5)(-0.1)$$ $$L(-0.1)=1-0.05$$ $$L(-0.1)=0.95$$

So this tells us that $$\sqrt{0.9} \approx 0.95$$.

## What if you aren’t given a function f(x) or an x=a value to use?

This seems nice to be able to estimate complicated values using this technique, but what if you need to estimate a complex value and aren’t given a function to use. Sometimes you aren’t given a function. And if you were trying to use this technique in the real world you probably wouldn’t be given a nice function to use and an a value that’s close to the input you need to estimate.

So how do you deal with this?

Well, the short answer is that you need to come up with the function and the a value on your own. Then once you have these you can apply the same technique above. Let me show you what I mean.

Let’s say you are told:

Use linear approximation to estimate $$\sqrt{24}$$.

The first thing you want to do is come up with the function to use to apply the linearization formula to. Since we are trying to find $$\sqrt{24}$$, our function is clearly going to need a square root in it somewhere. It is possible that you may need to experiment a bit and see how it works out, so let’s just start with the most obvious choice for now and say $$f(x)=\sqrt{x}$$.

If we do say $$f(x)=\sqrt{x}$$ then we can see that $$f(24)=\sqrt{24}$$ which is exactly the number that we are trying to estimate. But f(24) is hard to evaluate. So to find the a value we will want to use, you should consider what number is near 24 that we could plug into the function $$f(x)=\sqrt{x}$$ and easily figure out the result?

The closest number to 24 that we can easily plug into our function would be 25. Because 25 is a perfect square, we know that $$f(25)=\sqrt{25}=5$$, which is a nice round number. This is perfect for linear approximation because we need an input that is near the location we are trying to estimate whose output we know.

Since 25 is near 24 and we know the exact value of f(25), we can find the linear approximation of $$f(x)=\sqrt{x}$$ at a=25 and use this to estimate $$f(24)=\sqrt{24}$$ just like we did in the example above.

I’m not going to work this all the way through because it will be the exact same process as the last example now that we know $$f(x)=\sqrt{x}$$ and a=25. You can also see the rest of the solution to this problem in the video above.

## Differentials and how to find dy

Differentials go hand-in-hand with linear approximation and they have some interesting applications of their own.

Given some function f(x), you can find its differential, dy, simply by using the following formula.

$$dy=f'(x) \ dx$$

For example, say you are given the function $$y=e^{\frac{x}{10}}$$ and you are told to find the differential dy. All you need to do is apply the above formula.

You know that $$f(x)=e^{\frac{x}{10}}$$ Therefore, you can apply the chain rule to find $$f'(x)=e^{\frac{x}{10}} \cdot \frac{1}{10}$$ $$f'(x)=\frac{1}{10}e^{\frac{x}{10}}$$

So as a result, we know that

$$dy=\frac{1}{10}e^{\frac{x}{10}} \ dx$$

## Application of Differentials: Estimating Error

Estimating error is one of the more common uses for differentials. This can then be used to find the percentage error of a measurement. Let me explain what this means with an example.

Let’s say you measure a sphere to have a radius of 5 m. However, you know that the instrument you used to measure the radius of this sphere could have up to 5 cm (or 0.05 m) of error. Use differentials to estimate the maximum error in the measured volume of this sphere.

Knowing the the volume of a sphere is $$V=\frac{4}{3}\pi r^3$$ we will use this as our function. We will be able to use the differential of this function to estimate the maximum possible error in the measured volume of this sphere based on the maximum possible error in the radius. We already know the possible error in the radius, so we can use this and relate it to the volume.

First all we need to do is find the differential of our volume equation. This will require taking the derivative of $$\mathbf{V=\frac{4}{3}\pi r^3}$$. Using the formula of a differential as stated in the previous section, the differential is

$$dV=4 \pi r^2 \ dr$$

Then with this, all you need to do is plug in the information we know to find the possible error in the volume. Keep in mind, r represents the radius of the sphere that we know to be 5 m. And dr will represent the possible error in the radius, which we know to be 0.05 m. It’s important that we use the same units for all of these inputs since the output will use the same units in that case.

$$dV=4 \pi (5m)^2(0.05m)$$ $$dV=5 \pi m^3$$

So we know that the maximum error in the measurement we have for the volume of this sphere is $$\mathbf{5 \pi m^3\approx 15.708m^3}$$.

## Percentage Error

Once we find the maximum possible error as outlined above, we can use this to find the percentage error of our measurement. This is simply the maximum possible error in the measurement divided by the measurement itself. So in this case it would be

$$Percentage \ Error = \frac{error \ in \ volume}{volume}.$$

$$Percentage \ Error = \frac{5 \pi}{\frac{500}{3}\pi} = 3\%$$

## Limits to Infinity

Limits as x approaches infinity can be tricky to think about. This is because infinity is not a number that x can ever be equal to. To evaluate a limit as x goes to infinity, we cannot just simply plug infinity in for x and see what we get. As a result, things like $$\mathbf{e^{\infty}}$$ and $$\mathbf{\frac{1}{\infty}}$$ don’t actually have a value.

## So how can we deal with infinity?

Although infinity doesn’t have a specific value and can’t be plugged into functions, we can think about what will happen to a given function as x approaches infinity.

All this really means is that x is continually getting infinitely large. And as x gets bigger and bigger and bigger, what y value will our function get closer and closer to?

Let’s look at a few common examples and what they mean.

### One Divided by Infinity

Like I said before, infinity is not a value. Therefore, $$\frac{1}{\infty}$$ isn’t an actual number and doesn’t have a value. However, what we want to think about is what y value 1/x will approach as x goes to infinity. This is exactly what is being asked when we see: $$\lim_{x \to \infty} \frac{1}{x}$$

So let’s think about what happens to 1/x when we plug in bigger and bigger numbers for x.

So you can see in the table above that as x gets bigger and bigger, 1/x gets closer and closer to 0. Or in other words,

as x approaches infinity, 1/x approaches 0

We would write this mathematically as: $$\lim_{x \to \infty} \frac{1}{x} = 0$$

We can also see this graphically using Mathway. Notice in the graph below that as the x value goes toward infinity, you can see the y value getting closer to the y-axis (y=0).

### Limits Going to Infinity

The other common example I mentioned is the limit as x goes to infinity of $$\mathbf{e^x}$$. Or $$\lim_{x \to \infty} e^x$$

Again, it doesn’t really make sense to say that we can just plug infinity in for x and get $$\mathbf{e^{\infty}}$$. This doesn’t actually have a value. This isn’t a number. Instead, we want to think about what y value $$\mathbf{e^x}$$ goes toward as x goes to infinity. So let’s look at what happens as we raise e to a larger and larger power.

So we can see here that $$\mathbf{e^x}$$ starts giving us very large numbers quite quickly. And as we continue to plug in larger values for x, $$\mathbf{e^x}$$ will continue to get bigger and bigger and bigger.

as x approaches infinity, $$\mathbf{e^x}$$ approaches infinity

We would write this mathematically as: $$\lim_{x \to \infty} e^x = \infty$$

## Rational Functions

Finding the limit as x approaches infinity of rational functions is a common limit you will run into. This is important because this is how you find horizontal asymptotes of rational functions. You are just looking to see what y value your function will get really close to (without touching that value) as your x goes to infinity.

#### What is a rational function?

A rational function is a function that is a fraction where the top and bottom of the fraction are polynomials. Basically this just means that the numerator and denominator of the fraction will be a sum of a handful of terms that are a constant times x raised up to some power. So it will look like this: $$f(x)=\frac{a_nx^n + a_{n-1}x^{n-1} + … + a_2x^2 + a_1x + a_0}{b_mx^m + b_{m-1}x^{m-1} + … + b_2x^2 + b_1x + b_0}$$

### How do you take the limit of a rational function?

There are really only 3 cases you need to consider and the video above discusses these three cases as well. Any rational function will fall into one of these three categories, and each limit within each category will work out the same.

All you need to do is look at the degree of the polynomial on the top and bottom of the fraction. The degree of a polynomial is the highest power that x is being raised to. So for example, $$\mathbf{y=-4x^5+6x^2+x-12}$$ is a polynomial of degree 5, because the highest power of x is 5.

#### Case 1: Degree of numerator is larger than degree of denominator

If the degree of the numerator is higher than the degree of the polynomial on the denominator, then the limit will go to infinity or negative infinity. This will only depend on the sign of the coefficient of the highest power x term on the numerator.

If Degree(P(x)) > Degree(Q(x)), then $$\mathbf{\lim\limits_{x \to \infty} \frac{P(x)}{Q(x)}= \pm \infty}$$

#### Example:

$$\lim_{x \to \infty} \frac{x^4-3x^2+x}{x^3-x+2}$$ $$=\lim_{x \to \infty} \frac{x^4-3x^2+x}{x^3-x+2} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}}$$ $$=\lim_{x \to \infty} \frac{\frac{x^4}{x^3}-\frac{3x^2}{x^3}+\frac{x}{x^3}}{\frac{x^3}{x^3}-\frac{x}{x^3}+\frac{2}{x^3}}$$ $$=\lim_{x \to \infty} \frac{x-\frac{3}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}+\frac{2}{x^3}}$$ $$= \frac{\lim\limits_{x \to \infty}x-\lim\limits_{x \to \infty}\frac{3}{x}+\lim\limits_{x \to \infty}\frac{1}{x^2}}{\lim\limits_{x \to \infty}1-\lim\limits_{x \to \infty}\frac{1}{x^2}+\lim\limits_{x \to \infty}\frac{2}{x^3}}$$ $$= \frac{\lim\limits_{x \to \infty}x-0+0}{1-0+0}$$ $$=\frac{\lim\limits_{x \to \infty}x}{1}$$ $$=\lim_{x \to \infty}x$$ $$=\infty$$

#### Case 2: Degree of numerator is smaller than degree of denominator

If the degree of the numerator is smaller than the degree of the denominator then the limit will go to 0.

If Degree(P(x)) < Degree(Q(x)), then $$\mathbf{\lim\limits_{x \to \infty} \frac{P(x)}{Q(x)}= 0}$$

#### Example:

$$\lim_{x \to \infty} \frac{16x^4}{0.0001x^5+18x}$$ $$=\lim_{x \to \infty} \frac{16x^4}{0.0001x^5+18x} \cdot \frac{\frac{1}{x^5}}{\frac{1}{x^5}}$$ $$=\lim_{x \to \infty} \frac{\frac{16x^4}{x^5}}{\frac{0.0001x^5}{x^5}+\frac{18x}{x^5}}$$ $$=\lim_{x \to \infty} \frac{\frac{16}{x}}{0.0001+\frac{18}{x^4}}$$ $$= \frac{\lim\limits_{x \to \infty}\frac{16}{x}}{\lim\limits_{x \to \infty}0.0001+\lim\limits_{x \to \infty}\frac{18}{x^4}}$$ $$= \frac{0}{0.0001+0}$$ $$=0$$

#### Case 3: Degree of numerator is equal to degree of denominator

If the degree of the numerator is equal to the degree of the denominator then the limit will be equal to the coefficient of the highest power x term in the numerator divided by the coefficient of the highest power x term in the denominator.

If Degree(P(x)) = Degree(Q(x)), then $$\mathbf{\lim\limits_{x \to \infty} \frac{P(x)}{Q(x)}= \frac{a}{b}} \$$ where a is the coefficient of the highest power x term in P(x), and b is the coefficient of the highest power x term in Q(x).

$$\lim_{x \to \infty} \frac{x^2+7}{-3x^2}$$ $$=\lim_{x \to \infty} \frac{x^2+7}{-3x^2} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}$$ $$=\lim_{x \to \infty} \frac{\frac{x^2}{x^2} + \frac{7}{x^2}}{\frac{-3x^2}{x^2}}$$ $$=\lim_{x \to \infty} \frac{1 + \frac{7}{x^2}}{-3}$$ $$= \frac{\lim\limits_{x \to \infty}1 + \lim\limits_{x \to \infty}\frac{7}{x^2}}{\lim\limits_{x \to \infty}-3}$$ $$= \frac{1+0}{-3}$$ $$=-\frac{1}{3}$$

## Other Examples

$$\mathbf{\lim\limits_{x \to \infty} arctan \big( e^x \big)}$$ | Solution

$$\mathbf{\lim\limits_{x \to \infty} \frac{sin(x)}{x}}$$ | Solution

$$\mathbf{\lim\limits_{x \to \infty} \frac{ln(x)}{\sqrt{x}}}$$ | Solution

## Solutions

#### Example 1 Solution

$$\mathbf{\lim\limits_{x \to \infty} arctan \big( e^x \big)}$$

This is going to be based on the fact that we already discussed above that $$\mathbf{\lim\limits_{x \to \infty} e^x = \infty}$$. Since we know that, we can say that $$\mathbf{y=e^x}$$ and rewrite our limit as: $$\lim_{y \to \infty} arctan(y)$$

This is because y goes to infinity as x goes to infinity since $$\mathbf{y=e^x}$$. Now we can find this limit by looking at a graph of y=arctan(x).

Looking at this graph we can see that y=arctan(x) has a horizontal asymptote at $$\mathbf{y=\frac{\pi}{2}}$$. As x goes toward infinity, you can see that the y value of our function gets closer and close to $$\mathbf{\frac{\pi}{2}}$$. So this tells us $$\lim_{y \to \infty} arctan(y)=\frac{\pi}{2}$$ $$\lim_{x \to \infty} arctan \big( e^x \big)=\frac{\pi}{2}$$

#### Example 2 Solution

$$\mathbf{\lim\limits_{x \to \infty} \frac{sin(x)}{x}}$$

We know that $$\mathbf{-1 \leq sin(x) \leq 1}$$ for all x. Since we are looking at this limit as x goes to positive infinity, we can also say that $$-\frac{1}{x} \leq \frac{sin(x)}{x} \leq \frac{1}{x}$$

We also know that $$\mathbf{\lim\limits_{x \to \infty} -\frac{1}{x}=0}$$ and that $$\mathbf{\lim\limits_{x \to \infty} \frac{1}{x}=0}$$. Since we know $$\mathbf{\frac{sin(x)}{x}}$$ is between $$\mathbf{-\frac{1}{x}}$$ and $$\mathbf{\frac{1}{x}}$$ we can use Squeeze Theorem to say that $$\lim_{x \to \infty} \frac{sin(x)}{x}=0$$

#### Example 3 Solution

$$\mathbf{\lim\limits_{x \to \infty} \frac{ln(x)}{\sqrt{x}}}$$

For this limit, we will be able to use L’Hospital’s Rule. This is because ln(x) and $$\mathbf{\sqrt{x}}$$ both go to infinity as x goes to infinity. This gives us an indeterminate form that is a possible application of L’Hospital’s Rule. We can also check to make sure that this meets the other conditions needed to apply L’Hospital’s Rule.

$$\lim_{x \to \infty} \frac{ln(x)}{\sqrt{x}}$$ $$=\lim_{x \to \infty} \frac{\frac{d}{dx}ln(x)}{\frac{d}{dx}\sqrt{x}}$$ $$=\lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2x^{1/2}}}$$ $$=\lim_{x \to \infty} \frac{1}{x} \cdot \frac{2x^{1/2}}{1}$$ $$=\lim_{x \to \infty} \frac{2}{x^{1/2}}$$ $$=\lim_{x \to \infty} \frac{2}{\sqrt{x}}$$ $$=0$$

## Implicit Differentiation Examples

If you haven’t already read about implicit differentiation, you can read more about it here. Once you check that out, we’ll get into a few more examples below.

$$\mathbf{1. \ \ ycos(x) = x^2 + y^2}$$ | Solution

$$\mathbf{2. \ \ xy=x-y}$$ | Solution

$$\mathbf{3. \ \ x^2-4xy+y^2=4}$$ | Solution

$$\mathbf{4. \ \ \sqrt{x+y}=x^4+y^4}$$ | Solution

$$\mathbf{5. \ \ e^{x^2y}=x+y}$$ | Solution

For each of the above equations, we want to find dy/dx by implicit differentiation. In general a problem like this is going to follow the same general outline. Although, this outline won’t apply to every problem where you need to find dy/dx, this is the most common, and generally a good place to start. Start with these steps, and if they don’t get you any closer to finding dy/dx, you can try something else. Here are the steps:

• Take the derivative of both sides of the equation with respect to x.
• Separate all of the dy/dx terms from the non-dy/dx terms.
• Factor out the dy/dx.
• Isolate dy/dx.

Some of these examples will be using product rule and chain rule to find dy/dx.

## Solutions

$$\mathbf{1. \ \ ycos(x) = x^2 + y^2}$$

$$ycos(x)=x^2+y^2$$ $$\frac{d}{dx} \big[ ycos(x) \big] = \frac{d}{dx} \big[ x^2 + y^2 \big]$$ $$\frac{dy}{dx}cos(x) + y \big( -sin(x) \big) = 2x + 2y \frac{dy}{dx}$$ $$\frac{dy}{dx}cos(x) – y sin(x) = 2x + 2y \frac{dy}{dx}$$ $$\frac{dy}{dx}cos(x) -2y \frac{dy}{dx} = 2x + ysin(x)$$ $$\frac{dy}{dx} \big[ cos(x) -2y \big] = 2x + ysin(x)$$ $$\frac{dy}{dx} = \frac{2x + ysin(x)}{cos(x) -2y}$$

$$\mathbf{2. \ \ xy=x-y}$$

$$xy = x-y$$ $$\frac{d}{dx} \big[ xy \big] = \frac{d}{dx} \big[ x-y \big]$$ $$1 \cdot y + x \frac{dy}{dx} = 1-\frac{dy}{dx}$$ $$y+x \frac{dy}{dx} = 1 – \frac{dy}{dx}$$ $$x \frac{dy}{dx} + \frac{dy}{dx} = 1-y$$ $$\frac{dy}{dx} \big[ x+1 \big] = 1-y$$ $$\frac{dy}{dx} = \frac{1-y}{x+1}$$

$$\mathbf{3. \ \ x^2-4xy+y^2=4}$$

$$x^2-4xy+y^2=4$$ $$\frac{d}{dx} \big[ x^2-4xy+y^2 \big] = \frac{d}{dx} \big[ 4 \big]$$ $$2x \ – \bigg[ 4x \frac{dy}{dx} + 4y \bigg] + 2y \frac{dy}{dx} = 0$$ $$2x \ – 4x \frac{dy}{dx} – 4y + 2y \frac{dy}{dx} = 0$$ $$-4x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x+4y$$ $$\frac{dy}{dx} \big[ -4x+2y \big] = -2x+4y$$ $$\frac{dy}{dx}=\frac{-2x+4y}{-4x+2y}$$ $$\frac{dy}{dx}=\frac{-x+2y}{-2x+y}$$

$$\mathbf{4. \ \ \sqrt{x+y}=x^4+y^4}$$

$$\sqrt{x+y}=x^4+y^4$$ $$\big( x+y \big)^{\frac{1}{2}}=x^4+y^4$$ $$\frac{d}{dx} \bigg[ \big( x+y \big)^{\frac{1}{2}}\bigg] = \frac{d}{dx}\bigg[x^4+y^4 \bigg]$$ $$\frac{1}{2} \big( x+y \big) ^{-\frac{1}{2}} \bigg( 1+\frac{dy}{dx} \bigg)=4x^3+4y^3\frac{dy}{dx}$$ $$\frac{1}{2} \cdot \frac{1}{\sqrt{x+y}} \cdot \frac{1+\frac{dy}{dx}}{1} = 4x^3+4y^3\frac{dy}{dx}$$ $$\frac{1+\frac{dy}{dx}}{2 \sqrt{x+y}}= 4x^3+4y^3\frac{dy}{dx}$$ $$1+\frac{dy}{dx}= \bigg[ 4x^3+4y^3\frac{dy}{dx} \bigg] \cdot 2 \sqrt{x+y}$$ $$1+\frac{dy}{dx}= 8x^3 \sqrt{x+y} + 8y^3 \frac{dy}{dx} \sqrt{x+y}$$ $$\frac{dy}{dx} \ – \ 8y^3 \frac{dy}{dx} \sqrt{x+y}= 8x^3 \sqrt{x+y} \ – \ 1$$ $$\frac{dy}{dx} \bigg[ 1 \ – \ 8y^3 \sqrt{x+y} \bigg]= 8x^3 \sqrt{x+y} \ – \ 1$$ $$\frac{dy}{dx}= \frac{8x^3 \sqrt{x+y} \ – \ 1}{1 \ – \ 8y^3 \sqrt{x+y}}$$

$$\mathbf{5. \ \ e^{x^2y}=x+y}$$

$$e^{x^2y}=x+y$$ $$\frac{d}{dx} \Big[ e^{x^2y} \Big] = \frac{d}{dx} \big[ x+y \big]$$ $$e^{x^2y} \bigg( 2xy + x^2 \frac{dy}{dx} \bigg) = 1 + \frac{dy}{dx}$$ $$2xye^{x^2y} + x^2e^{x^2y} \frac{dy}{dx} = 1+ \frac{dy}{dx}$$ $$x^2e^{x^2y} \frac{dy}{dx} \ – \ \frac{dy}{dx} = 1 \ – \ 2xye^{x^2y}$$ $$\frac{dy}{dx} \big(x^2e^{x^2y} \ – \ 1 \big) = 1 \ – \ 2xye^{x^2y}$$ $$\frac{dy}{dx} = \frac{1 \ – \ 2xye^{x^2y}}{x^2e^{x^2y} \ – \ 1}$$

The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?

This is a fairly common example of a related rates problem and a common application of derivatives and implicit differentiation. I’m sure you may have come across a related rates ladder problem like this. If I can offer one piece of advice for this type of problem it’d be this: don’t use this ladder, it always falls…

Alright, bad jokes aside, this is going to follow the same 4 steps as all the other related rates problems I’ve done. If you’d rather watch a video, then check out my video below. But otherwise, let’s jump into it with the usual process!

## 1. Draw a sketch

As always, we’ll start by drawing a quick sketch of all of the information that is being described in the problem. To do this we should first think about what information we have. First of all, we need to think about the shape that’s being formed with the ladder.

Since the ladder is standing on the ground and leaning up against a vertical wall, we can say that a triangle would be formed by the 3 objects in the problem. More specifically we know that the vertical wall forms a 90 degree angle with the ground. Therefore, the triangle formed by the ground, the wall, and the ladder would be a right triangle.

On top of this, the problem also gives us a few pieces of information about the dimensions of the triangle and how they are changing. It actually tells us about how fast the ladder is moving, but since the ladder is what forms the triangle, we can deduce how the dimensions of the triangle are changing.

We are given 3 pieces of information about the position of the ladder as well as how the ladder is moving at the specific instant we are looking at.

• Bottom of the ladder is 3 m away from the wall.
• Top of the ladder is moving down the wall at a rate of 0.15 m/s.
• Bottom of the ladder is moving away from the wall at a rate of 0.2 m/s.

Adding these labels to our drawing from above would give us something like this:

This sketch gives us a pretty good idea of what is going on in this problem. Not only that, but we will be able to use this to get an idea of what kind of equation we will need to come up with.

## 2. Come up with your equation

Before we come up with our equation we want to sort through the information we are given and asked to find. This is important because we need to decide what measurements and variables we want in the equation.

#### What are we looking for?

The question asks us to find the length of the ladder. Therefore, we will need to find the length of the hypotenuse of the triangle in our drawing. Because of this we want to be sure to include the hypotenuse of our triangle in our equation.

#### What do we know about?

Looking back up at our labeled drawing, you can see that we really only have information about the bottom and side of our triangle. We know the length of the bottom side of the triangle and the rate of change of this side.

And we were also given information about the rate of change of the left side of the triangle. But remember that our equation in this step cannot include rates of change. Instead, the fact that we know this rate of change tells us that we can use the left side length of the triangle in our equation, not its rate of change.

We aren’t given any information about the angles in the triangle other than the fact that it’s a right triangle. As a result, we probably don’t want our equation to involve the angles of this triangle.

Since we know that our equation either needs to include, or can include the lengths of the sides of the triangle we should label them. Let’s go back to our drawing and label the side of the triangle. You can label them whatever you’d like, but I’ll go with x, y, and z.

#### Putting it into an equation

At this point we’ve figured out that we need an equation that related the sides of a right triangle.

What do you know about the relationship between the sides of a right triangle, but neither of the other two angles?

You’re probably thinking Pythagorean Theorem. If you are, you’re right! Pythagorean Theorem tells us that the square of the hypotenuse is equal to the sum of the squares of the other two sides of a right triangle. Remember this can only be applied to the sides of a right triangle, so noticing that is actually very important. In other words we know $$z^2=x^2+y^2.$$

## 3. Implicit differentiation

Now that we have come up with our equation, we need to apply implicit differentiation to take the derivative of both sides of our equation with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$

Before we do this though I want to point something out. Let’s look at each of the letters in this equation and consider how we need to treat them when we differentiate with respect to time.

Consider z first. z represents the length of the hypotenuse of the triangle. This is the side that is formed by the ladder. As time changes what happens to the length of the ladder? Nothing. It doesn’t change at all. It’s constant. Therefore we can treat z like a constant. If z is a constant and never changes, then $$z^2$$ would be constant too. It doesn’t change as time changes.

So when we take the derivative of z with respect to time, the derivative will be 0. The derivative of any constant is 0.

Unfortunately x and y won’t be as convenient. Looking back at our drawings you can see that the sides labelled x and y are changing over time. As the ladder slides down and away from the wall, these two sides of the triangle change in length. Therefore, when we take the derivative of $$x^2$$ and $$y^2$$ we will need to treat x and y as functions of time. Doing this means that we will need to use the chain rule, where x and y are the inside function and they are being plugged into another function that squares them.

#### Back to the derivative

Knowing how to treat each letter in our equation, let’s go ahead and take the derivative of both sides with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$ $$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$

## 4. Solve for the desired rate of change

Now all we need to do is plug in all of the information we have and solve for the right variable. However, this one is a little weird. The reason I say this is that we are actually not looking for a rate of change.

Remember the question told us to find the length of the ladder. Which means we need to find the value of z. The differential equation we just ended up with doesn’t even have a z in it, so how can we use it to find z?

Well, we’re actually going to need to go back to our original equation. $$z^2=x^2+y^2$$ We know the value of x based on information we were given, but we don’t know the value of y yet. If we could figure out what y was, then we could use this equation, plug in the value for x and y, then solve for z.

#### How do we find y?

This is what we will use our differential equation from the previous step for. That equation has a y in it, and we know the value of all the other variables.

• We are told that the moment we are considering is when the bottom of the ladder is 3 m from the wall. Since that corresponds to the side of our triangle labelled x, we know $$\mathbf{x=3}$$.
• We are also told that the ladder is moving away from the wall at a rate of 0.2 m/s. Therefore, x must be getting longer, or increasing, at that rate. So $$\mathbf{\frac{dx}{dt}=0.2}$$.
• And finally, the ladder is sliding down the wall at a rate of 0.15 m/s. So y must be getting shorter, or decreasing, at that rate. This means $$\mathbf{\frac{dy}{dt}=-0.15}$$.

#### Plugging it all into our equation

Knowing all of the values in our equation aside from y, we can plug these in and solve for y.

$$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$ $$0 = 2(3)(0.2) + 2y(-0.15)$$ $$0=1.2-0.3y$$ $$0.3y=1.2$$ $$y=4$$

Now that we know x and y, we can plug them back into our original equation and solve for z.

$$z^2=x^2+y^2$$ $$z^2=(3)^2+(4)^2$$ $$z^2=25$$ $$z=5$$

So the ladder must be 5 m long!