I've been trying to solve this limit, but can't really seem to be able to without using l'Hospital's rule (which the textbook specifically forbids). Here goes: $$\lim_{x\to0}\frac{\ln\left(\frac{e^x}{x+1}\right)}{x^2}\;.$$
I've messed up once already, since I tried to use arithmetic of limits, ended up getting $0$ as the answer (which is supposed to be $0.5$), only to realise that I can't use arithmetic of limits for this problem. Any ideas?
Expand the top as $$\ln (\frac{e^x}{1+x}) =\ln(e^x)-\ln(1+x)=x-\ln(1+x)$$ Then look at the Maclaurin expansion for $\ln(1+x)$, and you'll notice that the $x$ terms cancel and there is a $x^2$ term that cancels with the denominator yielding your limit.