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.