limits - Solving $lim_{xto0} x * ln x$

I needed to solve

$\lim_{x\to0} x * ln x$

and I wasn't sure how I would do it so I looked up the answer.

They used L'Hoptial to solve this and I don't understand why this works.

$\lim_{x\to0} x * ln x = \lim_{x\to0} \frac{ln x}{1/x} $ but I can't use L'Hopital here because this is

$\frac{undefined}{0}$, so I looked up if ln 0 is really undefined and it turns out that the limit of $ln 0$ is $-infinity$

My textbook says I can only use L'Hopital with $\frac{inf}{inf}$ or $\frac{0}{0}$, so why am I allowed to use L'Hopital in this case?

Answer

You can use L'Hopital if you rewrite your expression $$\frac{\ln x}{1/x}$$

As $x \to 0^+,$ we have the indeterminate form of $\frac{\infty}{\infty}$, so you are now licensed to take the derivative of the numerator and of the denominator and evaluate the limit.

$$\lim_{x\to0^+}\frac{\dfrac1x}{-\dfrac1{x^2}}
\;=\;-\lim_{x\to 0^+}x = 0$$