Evaluating $\lim_{x\rightarrow 0} \frac{\ln(1-x)+\sin(x)}{x^2 e^x} $ I got this exercise which I quite frankly can't wrap my head around
$$\lim_{x\rightarrow 0} \frac{\ln(1-x)+\sin(x)}{x^2 e^x} $$
The result should be:$\ -\frac{1}{2} $
I tried by derivating the whole function and that led me nowhere. If I put in $\ x = 0 $, I get $\ \frac{0}{0} $ which means I can apply L'Hôpital's rule but that led me to the wrong answer as well. Obviously I'm doing something wrong. I had a look at Wolfram Alphas step-by-step solution to this exercise but I didn't really understand it.
I'm appreciative with any possible help.
Thanks, 
Michael.

Using L'Hospital's rule seems like a good way to go. 
Taking the derivative of the top and bottom you get $$\frac{ - \frac{1}{1-x} + \cos(x)}{2x e^x + x^2 e^x}$$
This is still $\frac{0}{0}$ so you can use L'Hospital again to get $$ \frac{-\frac{1}{(1-x)^2} - \sin(x)}{2e^x + 2xe^x + 2xe^x + x^2 e^x}$$
which evaluates to $-\frac{1}{2}$
