I needed to solve

$$\lim_{x \to 0} x * \ln x.$$

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

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

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

  • $\begingroup$ The answer to the question asked is this: $\lim_{x\to 0+}1/x$ is $+\infty$, not $0$. $\endgroup$
    – vadim123
    May 6, 2014 at 15:25
  • $\begingroup$ By that logic, 1/0 is also undefined and we can never use L'Hospital $\endgroup$
    – evil999man
    May 6, 2014 at 15:58

2 Answers 2


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.

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


We can apply L'Hospital's Rule safely on $$F=\lim_{x\to0^+}\frac{\ln x}{\dfrac1x}$$ as it is of the form $\dfrac\infty\infty$

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .