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?

  • $\begingroup$ 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 $a>0$. $\endgroup$ – ms_ Feb 3 '20 at 12:24
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    $\begingroup$ From that Wikipedia page: “Let I be an open interval containing c ... or an open interval with endpoint c (for a one-sided limit)” $\endgroup$ – 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.


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