Analyzing unbounded limits: mixed function | Limits and continuity | AP Calculus AB | Khan Academy

So, we're told that ( f(x) ) is equal to ( \frac{x}{1 - \cos(e^x) - 2} ), and they ask us to select the correct description of the one-sided limits of ( f ) at ( x = 2 ).

We see that right at ( x = 2 ), if we try to evaluate ( f(2) ), we get ( \frac{2}{1 - \cos(2) - 2} ), which is the same thing as ( \cos(0) ). Cosine of 0 is just 1, so ( 1 - 1 ) is 0. Therefore, the function is not defined at ( x = 2 ). That's why it might be interesting to find the limit as ( x ) approaches 2, especially the one-sided limits.

If the one-sided limits... well, I'll just leave it at that. So let's try to approach this. There are actually a couple of ways you could do it. One way is to do this without a calculator, by just inspecting what's going on here and thinking about the properties of the cosine function. If that inspires you, pause the video and work it out; I will do that at the end of this video.

The other way, if you have a calculator, is to do it with a little bit of a table like we've done in other example problems. If we think about ( x ) approaching 2 from the positive direction, we can make a little table here where you have ( x ) and then ( f(x) ). If we're approaching 2 from values greater than 2, you could have 2.1, 2, and 2.01.

Now, the reason why I said calculators is that these aren't trivial to evaluate because for ( 2.1 ), the evaluation becomes ( \frac{2.1}{1 - \cos(2.1) - 2} ). Evaluating it yields approximately 0.1. I do not know what cosine of 0.1 is without a calculator.

I do know that cosine of 0 is 1, so this is very close to 1 without getting to 1, and it's going to be less than 1. Cosine is never going to be greater than 1. The cosine function is bounded between -1 and 1, so this thing is going to be approaching 1, but it’s going to be less than 1. It definitely cannot be greater than 1.

That’s actually a good hint for how you can just explore the structure here. Now, regarding ( 2.01 ):

Well, that’s going to be ( \frac{2.01}{1 - \cos(0.01)} ), and this is going to be even closer to 1 without being 1. So this could... you know, this is... but it’s going to be less than 1, no matter what. Cosine of anything is going to be between -1 and 1, and it could even be including those values.

However, as we approach 2, this thing is going to approach... it’s going to approach 1. I guess you could say it approaches 1 from below.

You can start to make some intuitions here if it’s approaching 1 from below. This fraction here is going to be positive, and as we approach ( x = 2 ), well, the numerator is positive and approaching 2, while the denominator is positive. Therefore, this whole thing has to be approaching a positive value. Or, it could become unbounded in the positive direction, as we’ll see this is unbounded because this thing is even closer to 1 than this other expression.

You would see that if you have a calculator. But needless to say, this is going to be unbounded in the positive direction, heading towards positive infinity.

So, these two choices suggest that. We can make the same argument as we approach ( x ) from below or as we approach 2 from below, I should say. So, that is ( x ) and ( f(x) ).

Once again, I don’t have a calculator in front of me. You could evaluate these things with a calculator. It'll become very clear that these are positive values, and as we get closer to 2, they become larger and larger positive values.

The same thing would happen if you did 1.9, or even 1.99. Because here, you'd be looking at ( \frac{1.9}{1 - \cos(1.9) - 2} ). Now, here you’d have 1.9 over ( 1 - \cos(-0.1) ); while ( \cos(-0.1) ) is the same thing as ( \cos(0.1) ).

So, these two things are going to be equivalent, and once again we’ll be approaching positive infinity. The only choice where this holds is the first one. Whether we approach from the right-hand side or the left-hand side, we're approaching positive infinity.

Now, the other way you could have deduced that is to say, okay, as we approach 2, the numerator is going to be positive because 2 is positive. Then, over here, as we approach 2, cosine of anything can never be greater than 1; it’s going to approach 1 but be less than 1.

So, if this goes towards less than one as ( x ) approaches 2, then ( 1 - \text{something less than 1} ) is going to be positive. Therefore, you have a positive divided by a positive. You’re definitely going to get positive values as you approach 2.

And we know— or they’ve already told us— that these are going to be unbounded based on the choices. So, you would also pick that. But you should also feel confident about it. The closer that we get to 2, the closer this value right over here gets to 0.

And the closer we get to 1, the smaller the denominator gets. Then, dividing by smaller and smaller denominators will lead us to become unbounded towards infinity, which is exactly what we see in that first choice.