Undefined limits by direct substitution | Limits and continuity | AP Calculus AB | Khan Academy

Let's see if we can figure out the limit of x over natural log of x as x approaches one. Like always, pause this video and see if you can figure it out on your own.

Well, we know from our limit properties this is going to be the same thing as the limit as x approaches one of x over the limit, the limit as x approaches one of the natural log of x.

Now, this top limit, the one I have in magenta, this is pretty straightforward. If we had the graph of y equals x, that would be continuous everywhere; it's defined for all real numbers and it's continuous at all real numbers. So, if it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one.

So, this is just going to be one. We just put a one in for this x, so the numerator here would just evaluate to a one. Then the denominator, natural log of x, is not defined for all x's and therefore it isn't continuous everywhere. But it is continuous at x = 1.

Since it is continuous at x = 1, then the limit here is just going to be the natural log evaluated at x = 1. So this is just going to be the natural log, the natural log of one, which of course is zero.

e to the 0 power is 1, so this is all going to be equal to, this is going to be equal to, we just evaluate it: 1 over 0.

Now we face a bit of a conundrum. 1/0 is not defined. If it was 0 over 0, we wouldn't necessarily be done yet; that's an indeterminate form. As we will learn in the future, there are tools we can apply when we're trying to find limits and we evaluate it like this and we get 0 over 0.

But 1 over 0, this is undefined, which tells us that this limit does not exist. So, does not exist, and we are done.