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.

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

Finding Derivatives with Limits

At this point we should have at least a basic understanding of limits and how to find some limits.  However, I have only really discussed limits by themselves and not how they relate to the rest of calculus.  They are very important in calculus because they are used to define the most important calculus topics.

For example, the main topic which will be discussed for quite some time is derivatives.  Derivatives will come up in a lot of different settings, like finding rate of change, instantaneous rate of change, velocity, slope, and a few others.  The main thing to realize is that a derivative is generally used to find out how quickly, or slowly, something is changing.

I will go further into all of these things later, but for now I want to focus on the definition of derivatives and how to find a derivative using the definition.

The definition of a derivative

If we have some function, $f(x)$, we would write “the derivative of f” as $f'(x)$.  And we would define the derivative of f by using this limit:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$

This limit can be a bit confusing, so there’s something I would like to point out before we actually begin working with this limit.  The confusing thing here is that we have $x$ and $h$ in this limit and it looks as if they are both variables.  However, when we find this limit, we can only treat $h$ as a variable.  We will need to treat x as a constant and h as the only variable.

The reason for this is that we are finding the limit as $h$ goes to $0$.  This tells us that $h$ is moving in toward $0$.  It does not tell us that $x$ is changing at all.  Therefore, when we are working with the limit, we will act as if $x$ is a number, or a constant.  This means that once we find the limit, our answer may have $x$ in it still and this is completely fine since $x$ isn’t the variable in this case.  Now let’s try an example.

Example 1

Consider the function $f(x) = 4x^2 - 7x + 12$.  We will use the limit definition to find the derivative of this function, but first let’s break it down and consider each part on its own.

Finding f(x+h)

The fist thing we need to find is $f(x+h)$.  This notation basically just means that we need to look at our function $f$, and plug in $(x+h)$ wherever we see the input.  In other words, we need to replace all of the $x$‘s in the function with $(x+h)$‘s.  So,

$$f(x+h) = 4(x+h)^2 – 7(x+h) + 12.$$

Then we will want to expand this out so it’s easier to work with.  Remember $(x+h)^2$ is the same as $(x+h)(x+h)$, which means we need to foil it.

$$f(x+h) = 4(x+h)(x+h) – 7(x+h) + 12$$

$$=4(x^2 + xh + xh + h^2) – 7(x+h) + 12$$

$$=4(x^2 + 2xh+ h^2) – 7(x+h) + 12$$

$$=4x^2 + 8xh+ 4h^2 – 7x – 7h + 12$$

Since there aren’t any like terms we will leave it at that for now.

Putting it all together

Now we can put that into the rest of the equation.  Since we now know $f(x+h)$ and $f(x)$, we can plug those into the equation.  I would recommend surrounding each of them with a set of parenthesis so you don’t forget to distribute the negative sign in front of the $f(x)$.  This is a very common mistake, so be careful not to forget that because it will give you the wrong answer.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$

$$= \lim_{h \to 0} \frac{(4x^2 + 8xh+ 4h^2 – 7x – 7h + 12) – (4x^2 – 7x + 12)}{h}$$

Solving the limit

When I first see a limit, the first thing I usually consider is whether we can simply plug in $0$ for $h$. Essentially, I try to treat this function as if it were continuous at $h=0$ (remember $h$ is the variable here).

However, if we do this here we will get $0$ on the denominator.  Since you cannot divide by 0, this will not work.  So our strategy will be to simplify this fraction to a point where we can plug in $0$ for $h$.  The simplest way to do this is to rearrange the numerator so that we can cancel an $h$ from the numerator and denominator and get rid of our fraction all together.

$$f'(x)= \lim_{h \to 0} \frac{4x^2 + 8xh+ 4h^2 – 7x – 7h + 12 – 4x^2 + 7x – 12}{h}$$

$$= \lim_{h \to 0} \frac{8xh+ 4h^2 – 7h}{h}$$

At this point I would like to point something out. Notice, after simplifying the numerator of the fraction, each term remaining contains an $h$ in it.  This is important because it allows us to factor the $h$ out and cancel it with the $h$ in the denominator, getting rid of the fraction.  This will be an extremely common strategy to use for finding the derivative of a function using the limit definition.

$$f'(x)= \lim_{h \to 0} \frac{h(8x+ 4h – 7)}{h}$$

$$= \lim_{h \to 0} 8x+ 4h – 7$$

Now we have simplified to a point that we can solve this limit by plugging $0$ in for $h$.

$$f'(x)= 8x+ 4(0) – 7$$

$$= 8x – 7$$

So we have just shown that if $f(x)=4x^2-7x+12$, then $f'(x)=8x-7$.  We will later learn shortcuts like the product rule, quotient rule, and chain rule that will make finding derivatives like this much simpler and faster, but you will need to know how to find these using the limit definition.  The pattern shown in this problem is a common one.  It won’t work for all derivatives, but it’s a good thing to try first.

• It’s generally a good idea to see if you can reengage the top of the fraction in such a way that every term has $h$ as a factor.
• Then you can factor out the $h$, and cancel it with the $h$ on the bottom of the fraction.
• This usually leaves you with a function that you can directly plug $0$ into $h$ and simplify from there, leaving you with a function that doesn’t contain any $h$‘s, but usually contains $x$‘s.

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I would recommend checking out the other material I have about derivatives.  As I mentioned before, there are several shortcuts and methods that make this whole process a lot easier.  Go check out what I’ve written about on the derivatives page.  If you have a question that isn’t answered there just let me know by emailing me at jakesmathlessons@gmail.com.  I’ll do my best to answer any questions you send me and I may even post a lesson or full solution on it!

Continuity

Continuity is a relatively simple concept, but problems that require proving it can be a little tricky. Essentially, a continuous function is one that you can draw all in one motion without picking up your pencil. This is one explanation of what it means for a function to be continuous that I like because it doesn’t take any mathematical definitions or proofs to understand. Any holes or gaps in a function’s graph would be a discontinuity and would mean that the function is not continuous.

The limit definition of continuity

By definition, a function $f(x)$ is continuous at $x=a$ if $$\lim_{x \to a} f(x) = f(a).$$

Let’s think about what this equation is saying. The left side of this equation is something that we’ve already dealt with, limits. It’s asking us to find out what $y$ value we close in on as we travel along our function and close in on $x=a$. Keep in mind, $x$ is a variable here which represents the input of our function, $f(x)$, and $a$ is a constant. This means that $a$ represents some specific number, which could be any number.

The right side of the equation is simply asking us to plug that same $a$ value into $f(x)$ and take the $y$ value we get out.

In total, the above equation says that if we travel along a function and close in on a specific $x$ value, we should close in on the same $y$ value we would get if we simply plugged that $x$ value into the equation.

In other words, as we travel along a function toward a specific $x$ value, our $y$ value will also go toward the $y$ value of the function at that point. If that is the case, then the function is continuous at that specific point. If we can say that a function is continuous at every single possible value we could put in for $a$, then we can say that the function is continuous for all $x$. If this is true then we can draw our entire function in one motion without picking up our pencil!

Let’s do a couple examples.

Example 1

Remember back in the first lesson about limits, Limits – Intro, I said I would go back to discussing the importance of our limit in the first example giving us the same value as when we plugged $x=2$ into the equation?  I would like to go into that further.

The function we were considering was $f(x) = x^2$ and we were finding

$$\lim_{x \to 2} x^2.$$

After looking at the graph of this function, shown in Figure 1.1, we saw that

$$\lim_{x \to 2} x^2 = 4.$$

I also pointed out that plugging in $x=2$ directly into the function also returns a $y$ value of $4$. In other words, we know $$f(2) = 4.$$ Therefore, we know

$$\lim_{x \to 2} x^2 = 4 = f(2)$$

$$\lim_{x \to 2} x^2 = f(2).$$

Notice this is exactly like the definition of what it means for a function to be continuous at a point. If you replace $a$ with $2$, replace $f(x)$ with $x ^2$, and replace $f(a)$ with $4$, we have just shown that $y=x^2$ is a function that is continuous at $x=2.$

Example 2

The next example I want to discuss also goes back to a function we have already looked at. Referring back to Limits – Intro we will consider $f(x)$ shown in Figure 1.2.

The question we will answer here is whether this function is continuous at $x=1$ or not. What do you think?

There are a few different ways we can answer this question. The simplest would be to simply look at the graph of the function and think about whether we can draw that function at and around $x=1$ without picking up our pencil. As you can see, there is clearly a hole at $x=1$ where we would need to pick up our pencil, and add a single point at $(1, \ 4)$.

As a result, it is probably safe to say that this function is not continuous at $x=1$. However, we want to be able to show this using the actual definition of what it means for a function to be continuous.

Remember, for this function, which we are calling $f(x)$ in this case, we need to be able to show that $$\lim_{x \to 1} f(x) = f(1).$$

If we can show this equation to be true, then $f(x)$ is continuous at $x=1$, and if it’s not true then the function is not continuous at $x=1$. Luckily, back when I first used this function as an example in the Limits – Intro lesson, we found that $$\lim_{x \to 1} f(x) = 2.$$ Therefore, we just need to find out if $f(1)$ is also $2$ and we can prove that this function is continuous or not continuous.

Looking at the graph again, we see that this function has a hole at $x=1$ and includes the point $(1, \ 4)$. In other words, if we plug $1$ into $f(x)$ as our $x$ value, we get a $y$ value of $4$ out. This is the same as saying $f(1) = 4$.

Now, at this point we have figured out $$\lim_{x \to 1} f(x) = 2 \ and$$ $$f(1) = 4.$$

Therefore, $$\lim_{x \to 1} f(x) \neq f(1)$$ and we can say that $f(x)$ is not continuous at $x=1$.

More Examples

Find the values of $a$ and $b$ that make $f$ continuous everywhere.

$$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+3 & \mbox{if } 2\leq x<3 \\ 2x-a+b & \mbox{if } x\geq 3 \end{cases}$$

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There are also several other lessons and problems on my limits page.  It would be a good idea to get some practice with limits.  A lot of other more complex topics in calculus are based around limits so they are important to understand.  If you can’t find the topic you want to read about just let me know by emailing me at jakesmathlessons@gmail.com and I’ll do my best to answer your question!

Squeeze Theorem

The Squeeze Theorem is a useful tool for finding complex limits by comparing the limit to two much simpler limits. Squeeze Theorem tells us that if we know these three things:

$$1. \ \ \ g(x) \leq f(x) \leq h(x)$$

$$2. \ \ \ \lim_{x \to a} g(x) = L$$

$$3. \ \ \ \lim_{x \to a} h(x) = L$$

Then we also know that

$$\lim_{x \to a} f(x) = L$$

Keep in mind, requirement number 1 above only needs to be true around $x=a$. It does not need to be true at $x=a$ and it also does not need to be true when we get far away from $x=a$. This is because limits only care what is happening to a function when we get infinitely close to a given $x$ value, not at that $x$ value and certainly not far away from that $x$ value.

Example 1

You will typically see this used when you are finding the limit of a function which is made up of two functions being multiplied together and one of them is a trig function. For example, consider

$$\lim_{x \to 0}x ^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg).$$

At first glance, this limit looks terribly complicated. Especially because the function $f(x) = x ^4 sin \Big( \frac{1}{\sqrt{x}} \Big)$ doesn’t even exist at $x=0$ (plugging in $x=0$ directly causes you to divide by $0$, which can’t be done). However, if we use Squeeze Theorem, we can find this limit by finding two other limits that are much easier to find.

Squeeze Theorem says that we will need to find one function that is greater than or equal to $f(x)$ around $x=0$ (because we are finding the limit as $x \to 0$) and another function that is less than or equal to $f(x)$ around $x=0$. It is not always obvious which functions to use for comparison, but any function that is made up of $sine$ or $cosine$ of something multiplied by something else will usually follow the same pattern, so this is a good thing to try first.

What’s the pattern?

Consider the function $y = sin(x)$. No matter what you plug in for $x$ you will always get a $y$ value between $-1$ and $1$. This will be true regardless of what you are taking the $sine$ of. Even if you were to take $sin\Big( \frac{1}{\sqrt{x}} \Big)$. No matter what you put in for $x$ this will give you something between $-1$ and $1$, except if you try to put in $x=0$ (remember you can’t divide by $0$).

But remember, with limits it doesn’t matter what happens at the $x$ value we are approaching, only what is happening around that point. Therefore, it doesn’t matter that this function isn’t defined at $x=0$ because it is defined for all $x$ values near $x=0$.

The point I’m trying to make is that $sine$ of anything can be bound between $-1$ and $1$, so we can say:

$$-1 \leq sin(x) \leq 1, \ and$$

$$-1 \leq sin \bigg( \frac{1}{\sqrt{x}} \bigg) \leq 1$$

How does this relate to our example?

Now, the trick that makes this whole thing work is the fact that multiplying both sides of an inequality by a positive number maintains the inequality. Therefore, we can also multiply both sides of the inequality by a function that always outputs a positive number (or $0$).

For example, we can multiply both sides by $x^4$, or all three “sides” in this case. We can do this because any number raised to an even power will always be positive (or $0$). This gives us

$$x^4 \cdot \Bigg[ -1 \leq sin \bigg( \frac{1}{\sqrt{x}} \bigg) \leq 1 \Bigg]$$

$$x^4 \cdot (-1) \ \leq \ x^4 \cdot \Bigg(sin \bigg( \frac{1}{\sqrt{x}} \bigg) \Bigg) \ \leq \ x^4 \cdot (1)$$

$$-x^4 \ \leq \ x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) \ \leq \ x^4$$

Now, by using the Squeeze Theorem, since $x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg)$ is trapped between $-x^4$ and $x^4$ near $0$, if we find that if

$$\lim_{x \to 0} -x^4 = \lim_{x \to 0} x^4$$

then that will also be the answer for the limit we are looking for. Consider the graph of $y=-x^4$ and $y=x^4$ around $x=0$ shown below which was graphed using Desmos.

You can see that for both of these functions, as we travel along each function from both sides and get closer and closer to $x=0$, we get closer and closer to a $y$ value of $0$. This tells us that

$$\lim_{x \to 0} -x ^4 = 0, \ and$$

$$\lim_{x \to 0} x ^4 = 0$$

Putting it all together

These two facts in combination with the inequality we showed earlier: $$-x^4 \ \leq \ x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) \ \leq \ x^4,$$ tells us that

$$\lim_{x \to 0}x ^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) = 0.$$

Note: Compare these last 4 statements to the 4 statements at the beginning of this lesson.  The inequality is exactly the first thing I said we were looking to be able to show and the two limits are exactly like part 2 and 3.  So it’s no surprise that these three pieces combine to prove the limit we were trying to find.

Example 2

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

The Squeeze Theorem is an important application of limits in calculus, but I have written several other lessons and problem solutions about limits!  You should go check out my limits page to see what else I’ve done.  If I haven’t answered any questions you have on that page, let me know by sending me an email at jakesmathlessons@gmail.com and I’ll do my best to address your questions.

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Limits – Intro

Limits are an important topic to understand in calculus. The reason for this is that most of the other main categories in calculus are built around limits.

But what is a limit?

When you take the limit of a function, we are essentially asking: what $y$ value does this function approach as $x$ gets really really close to a certain value.

Now we’ll get into a few examples of what this looks like graphically. If you’d like to learn more about evaluating limits algebraically, I’d recommend starting with the 8 limit properties.

Example 1

For example, take the function graphed below, $y = x^2$. Imagine we travel along the function, from both sides, near $x=2$. As we get closer and closer to the $x$ value of $2$, what $y$ value do we also get closer and closer to?

We get closer and closer to a $y$ value of $4$. This is a simple example, but this is the general idea behind all limits. If you need to find a limit of a function as $x$ approaches a certain value, you can figure this out quite easily with a graph of the function.

Just trace along the function from both the left and right side of that $x$ value and move in towards the specific $x$ value. Whatever $y$ value you close in on is the value of the limit of that function as $x$ approaches the given value.

In this example, we would say: “the limit of $x^2$, as $x$ approaches $2$, is $4$.”  Or “the limit as $x$ approaches $2$, of $x^2$, is $4$.” In mathematical notation, that phrase would look something like this:

$$\lim_{x \to 2} x^2 = 4$$

It just so happens that the value of the limit is the same value we would get by plugging $x = 2$ into this function. What I mean by that is, $f(2) = 2^2 = 4$ is exactly what we got for the limit of $x^2$ as $x$ approaches $2$. This is not a coincidence. I will go into more detail about why this is important when we discuss continuous functions and continuity. But for now, let’s do a few harder examples.

Example 2

I’m going to stick with a couple more examples of finding limits using graphs, because it is a skill you will need to know, and I’ve noticed that it’s something a lot of introductory calculus students have a hard time with. So if finding limits from graphs is something that confuses you, don’t worry.

Let’s try finding a little more challenging limit. Below is a graph of $f(x)$.

Using the graph in Figure 1.2, we will consider two different limits. First let’s consider the following:

$$\lim_{x \to 1}f(x)$$

Let’s look at what’s going on in this graph at and around $x = 1$. It looks like the function was supposed to go through the point $(1,\ 2)$, but that point got taken out. Instead, there is a hole there, and the graph of $f(x)$ includes the point $(1,\ 4)$. Therefore, we can say $f(1) = 4$. All this means is that this function returns a value of $4$, or $y = 4$, when $x = 1$. However, this has no impact at the limit we are considering.

Limits aren’t impacted by what happens at a specific point, only by what is happening around that point. As a result, knowing that $f(1)=4$ doesn’t help us to find

$$\lim_{x \to 1}f(x).$$

So what do we need to look at?

To find this limit, we need to look at what’s happening around $x = 1$, and ignore what’s happening at $x = 1$. As we get closer and closer to $x = 1$, from the left and the right, what $y$ value do we get close to? Imagine traveling along this function, perhaps starting at $x = 2$, and moving to the left. As you pass $x = 1.5$, then $x = 1.25$, then $x = 1.125$, and so on, we get closer and closer to the hole I mentioned earlier. We get closer and closer to a $y$ value of $2$.

The same thing happens if we start on that function at $x = 0$ and move to the right, toward $x = 1$. From both sides we get closer and closer to $y = 2$.

It is important to realize that we never actually get to $y = 2$. This is because we never get to $x = 1$, just infinitely close to it. As I said before, when solving for a limit, it doesn’t matter what happens at the point it’s asking about. It only matters what happens as you get infinitely close. This is the reason why, in this example, $f(1) = 4$, but

$$\lim_{x \to 1}f(x) = 2.$$

The other important thing to point out in this example is that we approached $x = 1$ from both sides and they both led us closer and closer to the same $y$ value of $2$. If both sides didn’t give us the same value, this limit would not exist.  This is illustrated in greater detail in my lesson about one-sided limits.

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If you would like to continue reading about limits, check out my limits page.  There’s plenty more lessons and example problems to read there.  If your questions are not answered there, I’d love your feedback.  Email me at jakesmathlessons@gmail.com with any unanswered questions you have and I’ll make sure to answer your question!