Limit of $x\ln{x}$

Solution 1:

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