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## Determining limits using algebraic properties of limits: direct substitution

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# Undefined limits by direct substitution

AP Calc: LIM‑1 (EU), LIM‑1.D (LO), LIM‑1.D.1 (EK)

## Video transcript

- [Voiceover] Let's see if
we can figure out the limit of x over natural log of
x as x approaches one. And like always pause this
video and see if you can figure it out on your own. Well we know from out limit
properties this is going to be the same thing as the limit as x approaches one of x over 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 straight forward, 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 it's continuous to limit
as x approaches one of x. It's just gonna be this
evaluated x equals one. So this is just going to be one. We just put a one in for this x. For the numerator here we
just evaluate to a one. And then the denominator, natural log of x is not
defined for all x's, therefore it isn't continuous everywhere. But it is continuous at x equals one. And since it is continuous
at x equals one, then the limit here is just
gonna be the natural log evaluated at x equals one. So this is just going
to be the natural log the natural log of one. Which of course is zero. E to the zero power is one. So this is all going to be equal to this is going to be equal to we just evaluate it one over one over zero. And now we face a bit of a conundrum. One over zero is not defined. It is was zero over zero,
we wouldn't necessarily be done yet but it's 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 zero over zero. But one over zero. This is undefined which tells us that this limit does not exist. So does not exist. And we are done.

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