Limit of $x\ln x$ as $x$ approaches $0^+$ How could I solve the limit in this form  $ \frac{x}{x+1}$ using l'Hospital's rule?
I know how to solve it in this way: $\frac{\ln x}{\frac{1}{x}}$
Thanks

Rewrite it like this
$$\lim_{x\to 0^+} x\ln x = \lim_{x\to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0 $$
EDIT: Your question is vague. Do you want to turn it into $\frac{0}{0}$ instead of $\frac{\infty}{\infty}$? That's not possible, as egreg pointed out.
