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

You can try using l'Hôpital's theorem:
$$
\lim_{x\to0}x\ln x=
\lim_{x\to0}\frac{x}{1/\ln x}\overset{\text{(H)}}{=}
\lim_{x\to0}\frac{1}{(-1/(\ln x)^2)(1/x)}=
\lim_{x\to0}-x(\ln x)^2
$$
This goes nowhere, if you're adamant into transforming the expression into a limit of the form $0/0$: the next step will take you to
$$
\lim_{x\to0}\frac{1}{2}x(\ln x)^3
$$
and so on.
It's like being inside a well; you have two directions: down or up. Which one do you choose?
