Limits at infinity of quotients with trig (limit undefined) | AP Calculus AB | Khan Academy

Let's see if we can figure out what the limit of ( x^2 + 1 ) over ( \sin(x) ) is as ( x ) approaches infinity.

So let's just think about what's going on in the numerator and then think about what's going on in the denominator. In the numerator, we have ( x^2 + 1 ). As ( x ) gets larger and larger and larger, as it approaches infinity, we're just squaring it here. So this numerator is going to get even bigger and approach infinity even faster. Thus, this thing is going to go to infinity as ( x ) approaches infinity.

Now what's happening to the denominator here? Well, ( \sin(x) ) – we've seen this before. ( \sin(x) ) and ( \cos(x) ) are bounded. They oscillate between negative 1 and 1. So, negative 1 is going to be less than or equal to ( \sin(x) ), which will be less than or equal to 1. Therefore, this denominator is going to oscillate.

So what does that tell us? Well, we might be tempted to say that the numerator is unbounded and goes to infinity, and then the denominator is just oscillating between these values here. So maybe the whole thing goes to infinity. But we have to be careful because one, this denominator is going between positive and negative values.

So, the numerator is just going to get more and more positive, but we're being divided sometimes by positive values and sometimes by a negative value. We're going to jump between positive and negative, positive and negative.

Then you also have all these crazy asymptotes here. Every time ( x ), every time ( \sin(x) ) becomes zero, well then you're going to have a vertical asymptote. This thing will not be defined. So you're going to have all these vertical asymptotes. You're going to oscillate between positive and negative and just larger and larger values.

And so this limit does not exist. Does not exist. Does not exist.

We can see that graphically. We've described it in words just inspecting this expression, but we can see it graphically. If we actually look at a graph of this, which I have right here, you can see that as ( x ) goes towards positive infinity, depending on which ( x ) we are, we're kind of going up. We get really large, then we hit a vertical asymptote, and we jump back down to a really negative value. Then another vertical asymptote, up, down, up, down, up, down.

It just is the oscillations that get more and more extreme, but we keep having these vertical asymptotes on a periodic basis. So it's very clear that this limit does not exist.