Limit of $x\ln{x}$

I am trying to solve $$\lim \limits_{x \to 0}x\ln{x}$$ which according to WolframAlpha (and Wikipedia) equals $0$.

I managed to solve it by substituting such that $y = \dfrac{1}{x}$ and then using L'Hôpital's rule:

$$\begin{align} \lim \limits_{x \to 0}x\ln{x} & = \lim \limits_{y \to \infty}\frac{\ln{\frac{1}{y}}}{y} \\ & = \lim \limits_{y \to \infty}\frac{-\frac{1}{y^2}{\frac{1}{\frac{1}{y}}}}{1} \\ & = \lim \limits_{y \to \infty}\frac{-1}{y} \\ & = 0 \end{align}$$

but my question is when I try to solve it using L'Hôpital's rule without making the substitution, I get:

$$\begin{align} & \lim \limits_{x \to 0}x\ln{x} \\ & = \lim \limits_{x \to 0}\frac{x}{x}+\ln{x} \\ & = \lim \limits_{x \to 0}1+\ln{x} \\ & = -\infty \end{align}$$

So what went wrong here? Is it because I made $\frac{x}{x}=1$? If so how would I proceed from that point?

Or is this one of those cases where the caveat in L'Hôpital's rule that:

$$\lim \limits_{x \to c}\frac{f'(x)}{g'(x)}$$

has to exist is violated? Does equaling $-\infty$ count as not existing?

1 Answer

In short if the limit of $f$ and $g$ are both zero or both $\pm\infty$, and the limit $f'/g'$ exists, then the limit $f/g$ equals it.

What's wrong is the expression $x\ln x$ as you are implicitly defining $f$ and $g$ doesn't meet the hypothesis. However if we write it as

$$\frac{\ln x}{1/x}$$

we can use l'Hopital with $f(x) = \ln x$, $g(x) = 1/x$ as

• The limits $\lim_{x\to 0^+} \ln x = -\infty$ and $\lim_{x \to 0^+} \frac{1}{x} = \infty$; and
• The limit $\displaystyle \lim_{x\to 0^+} \frac{f'(x)}{g'(x)}$ exists as $\displaystyle \lim_{x\to 0^+} \frac{1/x}{-1/x^2} = \lim_{x\to 0^+} -x = 0$

Hence $\lim\limits_{x\to 0^+} x\ln x = 0$.