Find $\lim _{x \rightarrow 0} x \ln(x)$. Use l'Hopital's rule if it applies. $$\lim _{x \rightarrow 0} x \ln(x)$$
I would say the limit will equal 1+ ln(x) but x cannot equal 0 at this point, so what shall i do ?
 A: You need to turn it into something that you can use l'Hopital on - that is, 0/0 or $\infty/\infty$.
For example, $$\frac{\ln(x)}{1/x}$$
A: Let $u=-\ln(x)$ which changes the problem to
$$- \lim_{u\to \infty} \frac{u}{e^u}.$$
Can you see it now?
A: $$\lim _{x \rightarrow 0} x \ln(x)=\lim _{x \rightarrow 0} \frac {\ln(x)}{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{\frac{-1}{x^2}}=-\lim _{x \rightarrow 0}x=0 $$
A: without L'Hopital 
$\lim _{x \rightarrow 0} x \ln(x) = \lim_{x \rightarrow 0} \frac{\ln(x)}{1/x}$
then, $ y = 1/x $ and when $x \rightarrow 0$, we know  $y \rightarrow \infty$
then our limit is $\lim_{y \rightarrow \infty} \frac{\ln(1/y)}{y} = \lim_{y \rightarrow \infty} \frac{\ln(1)- \ln(y)}{y} = \lim_{y \rightarrow \infty} \frac{- \ln(y)}{y} = \lim_{y \rightarrow \infty} -\frac{1}{y} \ln(y)$
$\lim_{y \rightarrow \infty} -\ln(y ^ \frac{1}{y} )$ 
And We know $\lim_{y \rightarrow \infty} y ^ \frac{1}{y} = 1 $ , then our limit
$\lim_{y \rightarrow \infty} -\ln(y ^ \frac{1}{y} ) = -\ln(1) = 0$  
