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

1 Answer
Henry
Jun 29, 2017

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:

enter image source here

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.

Related questions
See all questions in Limits at Infinity and Horizontal Asymptotes
Impact of this question
91667 views around the world
You can reuse this answer
Creative Commons License