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

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 $x \to {0}^{+}$.

#### 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 $x \to {0}^{+}$, 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 $\setminus \frac{\infty}{\setminus} \infty$ or $\frac{0}{0}$, then you can differentiate both parts to evaluate the limit. We need to rewrite the question to do this:

${\lim}_{x \to {0}^{+}} x \ln x = {\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{x}} = {\lim}_{x \to {0}^{+}} - \frac{\frac{1}{x}}{\frac{1}{x} ^ 2} = {\lim}_{x \to {0}^{+}} - x = 0.$

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.