TITLE: How to calculate $\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}$?
QUESTION [7 upvotes]: I am new at computing limits with infinitesimals and I am having trouble solving this one:
$$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}$$
I tried to substitute by means of equivalent infinitesimals and I came up with this:
$$\lim_{x\to0}\frac{x-x}{x^2}$$ But I do not know how to continue. The result must be $-1$. Any help would be appreciated!

REPLY [3 votes]: $$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}=\lim_{x\to0} \ln(1-x^2)^{1/x^2}=\lim_{t\to\infty} -\ln(1-\dfrac1t)^{-t}=-1$$