Evaluating $\lim_{x\rightarrow 0} \frac{\ln(1-x)+\sin(x)}{x^2 e^x} $

The $e^x$ part safely has limit $1$, so we can forget about it and calculate $$\lim_{x\to 0}\frac{\ln(1-x)+\sin x}{x^2}.$$ Now L'Hospital's Rule works nicely, with smaller probability of error than if we keep the $e^x$ part.

I prefer a somewhat more "hands on" approach that uses the first few terms of the power series expansions of $\ln(1-x)$ and $\sin x$. Note that $\ln(1-x)=-x-\dfrac{x^2}{2}+o(x^2)$ while $\sin x=x+o(x^2)$. Add. We get $-\dfrac{x^2}{2}+o(x^2)$. Divide by $x^2$. So our expression is $-\dfrac{1}{2}+o(1)$, which is the desired result.

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 $-\frac{1}{2}$