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

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?

• The answer to the question asked is this: $\lim_{x\to 0+}1/x$ is $+\infty$, not $0$. – vadim123 May 6 '14 at 15:25
• By that logic, 1/0 is also undefined and we can never use L'Hospital – evil999man May 6 '14 at 15:58

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