Vtgyjfy

How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ 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:

$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$

Thus:

$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$

By substituting: $t=ln(1-x)$ in the original limit we get:

$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$

Note that
$$begin{eqnarray}
lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
end{eqnarray}$$Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
$$ 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 $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
$$
frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
$$Thus we have
$$
frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
$$ and
$$
lim_{xto 0^+}g(x) =frac{1}{2}.
$$

First you have
$$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$

Now just use Taylor's formula at order $2$:
$$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
$$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$