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 . 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 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 , 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 anymore. Since we have some other number divided by zero, that tells us that this limit will likely be either or .

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 as ? 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 .

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 , . 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 you can see that the function is positive (above the *x* axis) to the right of . 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 , . 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 and the right-sided limit is . 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.