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
 A: 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?
A: 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.
A: If $x = e^{-y}$,
we want
$\lim_{y \to \infty} ye^{-y}$.
$ye^{-y}
=\frac{y}{e^y}
$.
But,
for $y > 0$,
$e^y
>1+y+y^2/2
>y^2/2
$,
so
$\frac{y}{e^y}
<\frac{y}{y^2/2}
= \frac{2}{y}
\to 0
$
as $y \to \infty$.
In general,
from the power series
for $e^y$,
for $y > 0$,
$e^y > \frac{y^{n+1}}{(n+1)!}$
so
$\frac{y^n}{e^y}
<\frac{(n+1)!}{y}
\to 0$
as $y \to \infty$.
