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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 \(f(x)=sin(x)\).

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.

x\(\mathbf{\frac{1}{x}}\)
11
100.1
1000.01
1,0000.001
10,0000.0001
100,0000.00001
1,000,0000.000001
10,000,0000.0000001
100,000,0000.00000001

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).

limit as x goes to infinity of 1/x

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.

x\(\mathbf{e^x}\)
12.718
27.389
454.598
6403.429
82,980.958
1022,026.466
1002.688 * \(\mathbf{10^{43}}\)

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).

limit as x approaches infinity of arctan(x) or tan^{-1}(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

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

This one is going to use Squeeze Theorem. Click here to learn more about Squeeze Theorem and its required conditions if you don’t already know about them, then come back to this problem.

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

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

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

\(\mathbf{2. \ \ xy=x-y} \)

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

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

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

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

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

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

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

RELATED RATES LADDER PROBLEM

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.

Ladder leaning against a wall
Ladder leaning against a wall

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:

related rates ladder sliding away from wall
Ladder sliding down and away from a wall

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.

triangle from ladder with sides labelled
Right triangle labeled with sides 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!

LIMIT PROPERTIES – Examples of using the 8 properties

I’ve already talked a bit about limits and one-sided limits and how to evaluate them, especially using the graph of the functions. But most limits that you need to evaluate won’t come with a graph and may be challenging to graph. In cases like these, you will want to try applying the 8 basic limit properties.

Using the limit properties is the simplest way to evaluate limits. Therefore, applying limit properties should be a good starting place for most limits. These properties can be applied to two-sided and one-sided limits.

First I will go ahead and list the 8 limit properties then I will show you a handful of examples that show how to apply these limits. These are the same 8 limit properties that are listed on my calculus 1 study guide. If you haven’t already, click here to download my calculus 1 study guide so you can have these limit properties handy as you work through evaluating limits with them.

What are the 8 limit properties?

\(\mathbf{1. \ \ \lim\limits_{x \to a} c = c} \)

Taking the limit of a constant just results in that constant.

\(\mathbf{2. \ \ \lim\limits_{x \to a} x = a} \)

The limit of the variable alone will go toward the value that the variable is approaching as given in the limit. This is a result of the fact that \(y=x\) is a continuous function.

\(\mathbf{3. \ \ \lim\limits_{x \to a} \Big( cf(x) \Big) = c \cdot \lim\limits_{x \to a} f(x)} \)

Having a constant being multiplied by the entire function within the limit can be pulled out of the limit. This will allow you to evaluate the simpler function, then multiply the result by that constant after evaluating a slightly simpler limit.

\(\mathbf{4. \ \ \lim\limits_{x \to a} \Big( f(x) \pm g(x) \Big) = \lim\limits_{x \to a} f(x) \pm \lim\limits_{x \to a} g(x)} \)

The limit of a sum or difference can instead be written as the sum or difference of their individual limits.

\(\mathbf{5. \ \ \lim\limits_{x \to a} \Big( f(x) \cdot g(x) \Big) = \lim\limits_{x \to a} f(x) \cdot \lim\limits_{x \to a} g(x)} \)

The limit of a product can instead be written as the product of their individual limits.

\(\mathbf{6. \ \ \lim\limits_{x \to a} \Big( \frac{f(x)}{g(x)} \Big) = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}, \ \ if \ \lim\limits_{x \to a} g(x) \neq 0} \)

Taking the limit of a quotient can be rewritten as the quotient of those two limits. Just make sure that the limit of the denominator isn’t zero. If it is, then this will result in dividing by zero, which you can’t do.

\(\mathbf{7. \ \ \lim\limits_{x \to a} \Big( f(x) \Big)^n = \Big( \lim\limits_{x \to a} f(x) \Big)^n} \)

Taking the limit of some function raised to a constant power can be rewritten to evaluate the limit of the inner function then raise the result to that constant power.

\(\mathbf{8. \ \ \lim\limits_{x \to a} \Big( \sqrt[\leftroot{-3}\uproot{3}n]{f(x)} \Big) = \sqrt[\leftroot{-3}\uproot{3}n]{\lim\limits_{x \to a} f(x)}} \)

Similar to the last property, but the same can be done with a function that is within a root. This can be applied to any constant root (eg. square root, cube root, etc.)

How can these limit properties be applied to evaluate limits?

These 8 properties of limits can be used to simplify limits and break them down into smaller pieces. Each of these smaller pieces would be easier to deal with. Then once you evaluate these smaller, simpler limits you can put them all together.

We will go ahead and show how to apply these limit properties with some examples. To the right of each step in parenthesis, I will put a number corresponding to the property from above that was used to get to that step from the previous one. If multiple properties were applied at the same time I will list all properties used in that step in the parenthesis.

Example 1

$$\lim_{x \to 5} 6x^4 – 2x + 7$$ $$= \ \lim_{x \to 5} 6x^4 – \lim_{x \to 5} 2x + \lim_{x \to 5} 7 \ \ \ \ (4)$$ $$= \ 6 \lim_{x \to 5} x^4 – 2 \lim_{x \to 5} x + \lim_{x \to 5} 7 \ \ \ \ (3)$$ $$= \ 6 \Big( \lim_{x \to 5} x \Big)^4 – 2 \lim_{x \to 5} x + \lim_{x \to 5} 7 \ \ \ \ (7)$$ $$= \ 6 (5)^4 – 2(5) + 7 \ \ \ \ (1, \ 2)$$ $$= \ 3,747$$

Example 2

$$\lim_{x \to 2} \Big( (x+2) \sqrt[\leftroot{-1}\uproot{3}3]{x^2 + 7x} \Big)$$ $$= \ \lim_{x \to 2}(x+2) \cdot \lim_{x \to 2}\sqrt[\leftroot{-1}\uproot{3}3]{x^2 + 7x} \ \ \ \ (5)$$ $$= \ \Big( \lim_{x \to 2}x+ \lim_{x \to 2}2 \Big) \cdot \lim_{x \to 2}\sqrt[\leftroot{-1}\uproot{3}3]{x^2 + 7x} \ \ \ \ (4)$$ $$= \ (2+2) \cdot \lim_{x \to 2}\sqrt[\leftroot{-1}\uproot{3}3]{x^2 + 7x} \ \ \ \ (1, \ 2)$$ $$= \ 4 \lim_{x \to 2}\sqrt[\leftroot{-1}\uproot{3}3]{x^2 + 7x}$$ $$= \ 4 \sqrt[\leftroot{1}\uproot{3}3]{\lim_{x \to 2} \Big(x^2 + 7x \Big)} \ \ \ \ (8)$$ $$= \ 4 \sqrt[\leftroot{1}\uproot{3}3]{\lim_{x \to 2} x^2 + \lim_{x \to 2} 7x} \ \ \ \ (4)$$ $$= \ 4 \sqrt[\leftroot{1}\uproot{3}3]{\Big( \lim_{x \to 2} x \Big)^2 + 7 \lim_{x \to 2} x} \ \ \ \ (7, \ 3)$$ $$= \ 4 \sqrt[\leftroot{1}\uproot{3}3]{(2)^2 + 7(2)} \ \ \ \ (2)$$ $$= \ 4 \sqrt[\leftroot{-1}\uproot{1}3]{18}$$

Example 3

$$\lim_{x \to 4} \frac{x}{28}$$ $$= \ \frac{\lim\limits_{x \to 4} x}{\lim\limits_{x \to 4} 28} \ \ \ \ (6)$$ $$= \ \frac{4}{28} \ \ \ \ (1, \ 2)$$ $$= \ \frac{1}{7}$$

Conclusion

As you can see, each of these properties can be applied to fairly complex limits to break them down into smaller, simpler pieces. Each will usually end in applying one of the first two properties listed above to convert a limit into some number. And in the end, you will end up converting all of the limits into numbers. At that point, you will be able to manipulate everything with simple algebra to simplify your answer.

If you’d like to get your own copy of my FREE STUDY GUIDE, you can get yours by clicking here. And check out and subscribe to my YouTube Channel as well for video versions of other topics that I have posted lessons about as well.

L’HOSPITAL’S RULE – HOW TO – With Examples

L’Hospital’s Rule really just tells us one thing that makes evaluating certain limits a lot easier. Limits that meet 3 specific requirements can be made much simpler using L’Hospital’s Rule. First let me introduce L’Hospital’s Rule, then we can go over the 3 conditions that you need to check before you can apply it to any given limit.

What does L’Hospital’s Rule tell us?

To find a limit of a function that is a fraction, we can take the derivative of the top of the fraction and the derivative of the bottom of the fraction and make a new fraction out of the derivatives.

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Notice that we don’t use the quotient rule here. The reason for this is that we are taking the limit of this fraction and not taking the derivative of this fraction. The limit is the key piece that allows us to avoid the quotient rule and take the derivative of each piece of the fraction separately to create another limit that is equal to the original.

The hope is that the fraction resulting from the derivatives will be easier to evaluate than the original limit was.

How do we know when to use L’Hospital’s Rule?

Before we can apply L’Hospital’s rule to any given limit, we need to confirm that these three conditions are met:

  1. f(x) and g(x) are differentiable on some open interval that includes \(\mathbf{x=a}\). This will basically just mean that both the numerator and denominator are differentiable at \(x=a\).
  2. \(\mathbf{g'(x) \neq 0}\) near \(\mathbf{x=a}\). Note that it doesn’t matter if \(g'(x)=0\) AT \(x=a\) as long as you can pick some interval (as small as is necessary) around \(x=a\) where \(g'(x) \neq 0\) for all x‘s in that interval besides \(x=a\). You likely won’t need to worry about running into a function that you can’t pick a small enough interval around \(x=a\) to make this work.
  3. As x \(\rightarrow\) a, f(x) AND g(x) \(\mathbf{\rightarrow 0}\) — OR — f(x) AND g(x) \(\mathbf{\rightarrow \pm \infty}\)

That’s really all there is to it. Let’s jump into some practice problems and I will show you how to apply L’Hospitals Rule.

Example 1

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x}$$

If we tried to use limit properties to evaluate this limit, we would see that both the top and the bottom of this fraction go to either positive or negative infinity as x goes to infinity.

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} \rightarrow \frac{\infty}{- \infty}$$

So L’Hospital’s Rule might help…

Now we just need to confirm that the other two conditions are met.

\(f(x)=e^x\) is an exponential function and is differentiable everywhere, for any value of x. And \(g(x)=-x^2 + 1000x\) is also differentiable everywhere since it’s a polynomial. Since it’s differentiable everywhere, it is also differentiable for any infinitely large x value.

\(g'(x) = -2x+1000\) will go to \(-\infty\) as x approaches \(\infty\). \(g'(x) \neq 0\) for any infinitely large x value since it just continues to go to \(– \infty\).

So we know that this limit meets all 3 requirements needed to apply L’Hospital’s Rule.

Now we know we can apply L’Hospital’s Rule

Taking the derivative of the top and bottom of the fraction individually tells us that:

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} = \lim_{x \to \infty} \frac{e^x}{-2x+1000}$$

Now we can evaluate this new limit instead. But if we do this, we will notice that we will still end up in the same situation that we had before.

$$\lim_{x \to \infty} \frac{e^x}{-2x+1000} \rightarrow \frac{\infty}{- \infty}$$

But what if it didn’t really help make the limit easier?

Which puts us in a perfect situation to consider using L’Hospital’s Rule to evaluate this new limit as well. By the same logic as before, we can confirm that the first two conditions are met as well as x gets infinitely large. Since all 3 required conditions are met we can go ahead and apply L’Hospital’s Rule a second time.

$$\lim_{x \to \infty} \frac{e^x}{-2x+1000} = \lim_{x \to \infty} \frac{e^x}{-2}$$

Now we can simply use the basic limit properties to evaluate this last limit.

Now we have a much easier limit

$$\lim_{x \to \infty} \frac{e^x}{-2} = \ – \frac{1}{2} \lim_{x \to \infty} e^x = \ – \infty$$

So therefore,

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} = \ – \infty$$

A quick note on applying L’Hospital’s Rule twice

This is an interesting problem because it shows that you can apply L’Hospital’s Rule multiple times on the same problem. You just need to make sure that each time you apply it, the resulting limit still meets all 3 required conditions before applying it to the new limit. There is no limit to the number of times you can continue applying L’Hospital’s Rule over and over in the same problem as long as you are making sure that the limit you are applying it to meets all 3 conditions every time you apply it.

Example 2

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to 3} \frac{x-3}{27-x^3}$$

Again, if we think about what value the top and bottom of this fraction will go towards as x approaches 3, we would see that

$$\lim_{x \to 3} \frac{x-3}{27-x^3} \rightarrow \frac{0}{0}$$

Since we get another indeterminate form, which is $\frac{0}{0}$, we should consider using L’Hospital’s Rule to make this limit easier to evaluate.

So L’Hospital’s Rule might help…

First, we need to make sure that the other two conditions are met as well.

We can check that \(g'(x) = -3x^2\) doesn’t equal zero anywhere near \(x=3\). This is because \(g'(x) = -3x^2\) is continuous everywhere and the only place where \(g'(x)=-3x^2=0\) is when \(x=0\). As a result of these two things, we can pick some interval around \(x=3\) that doesn’t include \(x=0\) to satisfy condition #2.

Also, both f(x) and g(x) are polynomials and are therefore differentiable everywhere. So we know they will both be differentiable on any interval around \(x=3\).

Now we know we can apply L’Hospital’s Rule

Doing so by taking the derivative of the top and bottom of our fraction separately tells us that

$$\lim_{x \to 3} \frac{x-3}{27-x^3} = \lim_{x \to 3} \frac{1}{-3x^2}$$

And doing this gives us an easier limit to deal with. Now we can simply apply the limit properties to evaluate. Applying the limit properties tells us that:

$$\lim_{x \to 3} \frac{1}{-3x^2} = \frac{1}{-3 \Big( \lim_{x \to 3}x \Big) ^2}= \frac{1}{-3(3)^2} = \ – \frac{1}{27}$$

So therefore we know that:

$$\lim_{x \to 3} \frac{x-3}{27-x^3} = \ – \frac{1}{27}$$

Example 3

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to 0} \frac{|x|}{x^5+2x}$$

If we think about what value the numerator and denominator of this fraction will approach as x approaches 0 from both sides, we would see:

$$\lim_{x \to 0} \frac{|x|}{x^5+2x} \rightarrow \frac{0}{0}$$

Since we get an indeterminate form, which is $\frac{0}{0}$, we should consider using L’Hospital’s Rule to make this limit easier to evaluate.

So L’Hospital’s Rule might help…

First, we need to make sure that the other two conditions are met as well.

Upon checking condition #1 however, we run into a problem. Condition #1 requires that f(x) and g(x) are both differentiable on some interval containing \(x=0\), including \(x=0\).

But \(f(x) = |x|\) is not differentiable at \(x=0\). Therefore, we actually can’t apply L’Hospital’s Rule to evaluate this limit. I won’t go into the details here since we won’t be using L’Hospital’s Rule. But if you want to try evaluating this limit, I’d recommend considering both one-sided limits on their own and compare them to start. You can see a similar application here.

Example 4

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

Optimization Problems Part 2

In my last lesson, I introduced optimization problems and I discussed local extrema. You should check that out if you haven’t already. The next thing that I would like to discuss now is finding global maximums and minimums. The first step in finding a global maximum or minimum of a function is actually very similar to finding the local max and min values.

But once you know about the local maximums and minimums, how do you find the global extrema?

Finding the global extrema from the local extrema is really quite simple. And there really is only one way to find the global maximum and minimum values of a function. You just need to find a list of all possible x values where the global max or global min may occur. Then once you have created a list of all possibilities, you just plug them all into the original function.

Not the function’s derivative or the function’s second derivative. But the original function.

You will test for global extrema of f(x) using f(x), not f'(x) or f”(x). This is after you have your list of all possible locations where the global extrema could occur (which will require the use of f'(x)). But my point is that there wouldn’t be a first derivative test or a second derivative test with the global extrema like there was with finding local extrema.

So what does this look like in practice?

Let’s use an example. For example, let’s say that we are asked to find the global maximum and the global minimum of \(f(x)=2x^3-\frac{5}{2}x^2+x-1\) on the domain \(-4 \leq x \leq 5\).

Notice we are being asked to find the global extrema on a specific domain. This is important because a lot of functions either go to infinity or to negative infinity as x either gets infinitely large, infinitely small, or approaches some specific value. So if we were asked to find the global maximum of a function that goes to infinity as x goes to infinity, we wouldn’t be able to do this. There would be no maximum since the function only continues to grow.

So we know we will be limited to a specific domain.

As I said before, finding global extrema starts out exactly like finding local extrema. The first thing we need to do is find the critical values of our function. To do this, we just need to find its derivative and set \(f'(x)=0\) and solve for x.

$$f(x)=2x^3-\frac{5}{2}x^2+x-1$$ $$f'(x)=6x^2-5x+1$$ Then set $$6x^2-5x+1=0$$ and solve for x to find the critical values. To do this, we can factor the left side of the equation. $$(3x-1)(2x-1)=0$$ To solve this we can set each factor equal to zero individually. $$3x-1=0 \ \ \ \ and \ \ \ \ 2x-1=0$$ $$3x=1 \ \ \ \ and \ \ \ \ 2x=1$$ $$x=\frac{1}{3}, \ \frac{1}{2}$$

So now we know that this function has two critical values, \(x= \frac{1}{3}\) and \(x=\frac{1}{2}\). Now this is where things get different with a global max/min problem versus finding the local max/min. We also need to consider the endpoints of our given domain as critical values!

This will always be the case when we are looking for a global maximum or minimum. The problem asked us to find the global extrema on the domain \(-4 \leq x \leq 5\). Therefore, we will also say that \(x=-4\) and \(x=5\) will be treated as critical values that we need to test.

So how to we test our critical values?

The first thing that I would like to do is list out all of the x values we will be testing in one place. Remember, the list of values we need to test came from two places:

  1. Setting \(f'(x)=0\) and solving for x.
  2. Each of the endpoints of the domain on which we need to find the global maximums and minimums.

So in this case we’ll have four total x values that we need to test: $$x=-4, \ \frac{1}{3}, \ \frac{1}{2}, \ and \ 5$$ To test these points, all we need to do is plug each of the four points into f(x). Whichever on outputs the largest number will tell us the global maximum. Whichever outputs the smallest number will tell us the global minimum.

$$f(-4)= \ 2(-4)^3-\frac{5}{2}(-4)^2+(-4)-1 \ = -173$$ $$f \bigg( \frac{1}{3} \bigg) = \ 2 \bigg( \frac{1}{3} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{3} \bigg) ^2+ \bigg( \frac{1}{3} \bigg) -1 \ = -\frac{47}{54}$$ $$f \bigg( \frac{1}{2} \bigg) = \ 2 \bigg( \frac{1}{2} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{2} \bigg) ^2+ \bigg( \frac{1}{2} \bigg) -1 \ = -\frac{7}{8}$$ $$f(5)= \ 2(5)^3-\frac{5}{2}(5)^2+(5)-1 \ = \frac{383}{2}$$

So we can see that the smallest of these four numbers is -173 and the largest of them is \(\frac{383}{2}\). Therefore, the global maximum of f(x) on \(-4 \leq x \leq 5\) is \(\frac{383}{2}\) which occurs when \(x=5\). And the global minimum of f(x) on \(-4 \leq x \leq 5\) is -173 which occurs when \(x=-4\).

We can even graph our function using Desmos along with the critical points to make sure our answer makes sense. You can click on the link in the last sentence to see a larger version of the graph.

global extrema maximum minimum
\(f(x)=2x^3-\frac{5}{2}x^2+x-1\) on the domain \(-4 \leq x \leq 5\)

Extra practice

If you’d like some extra practice finding global maximums and minimums, here are a few examples you can work through on your own. A couple of these examples will require the use of the product rule and quotient rule.

For each of the following, find the global maximum and minimum of the given function on the given domain or explain why one doesn’t exist.

$$f(x)= 2x^4 + 5x^2 – 12x \ \ \textrm{on the domain} -1 \leq x \leq 2$$ $$g(x)= xe^x +6x^3-12 \ \ \textrm{on the domain} -3 \leq x \leq 0$$ $$h(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -2 \leq x \leq \frac{3}{2} $$ $$j(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -\frac{9}{10} \leq x \leq \frac{3}{2}$$

Hopefully all of this helps with global maximums and minimums, but as always I’d love to hear your questions if you have any. If you find that you get stuck as you’re working through some of these extra practice problems just let me know. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I will send you my bonus FREE calculus 1 study guide to help you survive calculus! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about derivatives as well.