Calculate: $\lim\limits_{x\to0^-} \frac1{\ln(1-x)}+\frac1x$ without LHR/Expansions 
How to calculate $$\lim\limits_{x\to0^-} \left(\frac1{\ln(1-x)}+\frac1x \right)$$ without using L'Hopital, expansions nor integration?

I found the answer:
Using the Mean value theorem on:
$f(x)=e^x-\frac{x^2}2-x-1$
We get:
$0\le\frac{e^x-x-1}{x^2}-\frac1 2\le \frac{e^x-x-1}{x}$
Thus:
$\lim\limits_{x\to0^-} \frac{e^x-x-1}{x^2} = \frac12$
By substituting: $t=\ln(1-x)$ in the original limit we get:
$\lim\limits_{t\to0^+} \frac{1-e^t+t}{t(1-e^t)} = \lim\limits_{t\to0^+} \frac{e^t-t-1}{t^2}.\frac{t}{e^t-1} = \frac12$

Note that
$$\begin{eqnarray}
\lim_{x\to0^-} \left(\frac1{\ln(1-x)}+\frac1x \right)&=&\lim_{x\to0^+}\frac{x-\ln(1+x)}{x\ln(1+x)} \\&= &\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}\cdot  \lim_{x\to0^+}\frac{x}{\ln(1+x)}\\&=&\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}.
\end{eqnarray}$$Now, let $$g(x) =\begin{cases}\frac{x-\ln(1+x)}{x},\quad x\neq 0\\ 0,\quad x=0\end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}=\lim_{x\to0^+}\frac{g(x)}{x}=\lim_{c\to0^+} g'(c) =\lim_{c\to 0^+} \frac{\frac{c^2}{1+c}-(c-\ln(1+c))}{c^2}=1-L.
$$ This gives $L=\frac{1}{2}$.
(Justification of taking the limit) Let $h(x) = \ln(1+x) - x +\frac{x^2}{2}$. We have $h(x) \ge 0$ since $h(0) = 0$ and $h'(x) = \frac{1}{1+x}-1+x\ge 0$. This shows $\frac{g(x)}{x}\le\frac{1}{2}$. By MVT, we know that for some $c\in (0,x)$,
$$
\frac{1}{1+x}\le \frac{1}{1+c}= \frac{g(c)}{c}+\frac{g(x)}{x} .
$$Thus we have
$$
\frac{1}{1+x}-\frac{1}{2}\le g(x) \le \frac{1}{2},
$$ and
$$
\lim_{x\to 0^+}g(x) =\frac{1}{2}.
$$
