TITLE: Find $\lim\limits_{x\to 0}{\frac{x+\ln (1-x)}{x^2}}$ without derivatives
QUESTION [5 upvotes]: If we know that $$\lim\limits_{x\to 1}{\frac{\ln x}{x-1}}=1$$ and $$\lim\limits_{x\to 0}{\frac{x+\ln (1-x)}{x^2}}=l\in\mathbb{R}$$ find the value of $l$ without derivatives.
I have only managed to find that: $$\lim\limits_{x\to 0}{\frac{\ln (1-x)}{x}}=-1$$ Any hint?

REPLY [12 votes]: Change $x\to 0$ to $(-x)\to 0$ to get
$$
\lim_{x\to 0}\frac{-x+\ln(1+x)}{x^2}=l.
$$
Add two limits
$$
2l=l+l=\lim_{x\to 0}\Bigl(\frac{x+\ln(1-x)}{x^2}+\frac{-x+\ln(1+x)}{x^2}\Bigr)=
\lim_{x\to 0}\frac{\ln(1-x^2)}{x^2}.
$$
Change $x^2=t\to 0$ to get
$$
2l=\lim_{t\to 0}\frac{\ln(1-t)}{t}=-1.
$$
Thus, $l=-\frac12$.