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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:

$\lim\limits_{t\to0^+} \frac{1-e^t+t}{t(1-e^t)} = \lim\limits_{t\to0^+} \frac{e^t-t-1}{t^2}.\frac{t}{e^t-1} = \frac12$

Bilal HIMITE
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  • Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools? –  Jan 05 '19 at 15:52
  • Use https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Jan 05 '19 at 16:26

2 Answers2

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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 $$ \lim_{x\to 0^+}g(x) =\frac{1}{2}. $$

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

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 $$\frac1{\ln(1-x)}+\frac1x\sim_0\frac{-\cfrac{x^2}2}{-x^2} =\frac12.$$

Bernard
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