# Applying L'Hopital's rule in $\lim_{x \to 0^+} x \ln x$

I'm trying to evaluate $$\lim_{x \to 0^+} x \ln x$$, and here's what I got so far: $$\lim_{x \to 0^+} x \ln x \\ = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}}$$ and I wanted to apply L'Hopitals rule for that. However, the Wikipedia page of LH rule https://en.wikipedia.org/wiki/L%27Hôpital%27s_rule says that for the rule to work, both functions on the numerator and the denominator must be defined on a open interval containing the limit point, in this case, $$0$$. But since $$\ln x$$ isn't defined for negative x, does the rule really work in my case?

• Notice you are approaching the limit of $x$ approaching $0$ from the right (denoted by $\lim_{x\to 0^+}$. Thus this special case only requires the functions to be defined on a some interval $(0,a)$ for some $a>0$. – ms_ Feb 3 '20 at 12:24
• From that Wikipedia page: “Let I be an open interval containing c ... or an open interval with endpoint c (for a one-sided limit)” – Martin R Feb 3 '20 at 12:27

Since you are interested in the limit $$\lim_{x\to0^{\color{red}+}}x\ln x$$ here, then your function only has to be defined on some interval $$(0,a)$$ for some $$a>0$$. Since it is in fact defined in $$(0,\infty)$$, you have no problem here.