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:

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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.

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