Question

How do you find the limit of `xlnx` as `x->0^-`?

Answer 1

There is no limit as `x` approaches `0` from below since `ln x` is undefined for negative numbers. Instead, I will demonstrate how to find the right-handed limit, i.e., as `x->0^+`.

Explanation:

Here is a graph:

So we should expect the answer to be zero. Now, to do this we can't use the product rule, since the limit of `ln x` diverges as `x->0^+`, we have to be more clever.

L'HOPITAL'S RULE: You can Google the precise formulation of this, and the conditions where it applies, but roughly speaking, the rule states that if you have a limit of the form `\infty/\infty` or `0/0`, then you can differentiate both parts to evaluate the limit. We need to rewrite the question to do this:

`lim_{x->0^+}x ln x=lim_{x->0^+}ln x / {1/x}=lim_{x->0^+}-{1/x}/{1/x^2}=lim_{x->0^+}-x=0.`

You could probably figure out other ways to evaluate this limit, maybe using the squeeze theorem with upper bound `x^2` and something else for your lower bound, but L'Hopital's rule is how everyone would evaluate this limit.