Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Evaluate $\displaystyle \lim_{x \to 0} \left(\frac{1}{\ln(1+x)} + \frac{1}{\ln(1-x)}\right)$

I am having trouble starting this one. I couldn't see any log laws that I'm familiar with to rearrange the formula. I also tried combining the fractions and using L'Hospital's, but it only seemed to make things worse.

What direction should I take with this?

share|cite|improve this question
up vote 7 down vote accepted

What's wrong with L'Hopital's?

\begin{align*} \lim_{x\to 0}\frac{1}{\ln(1+x)}+\frac{1}{\ln(1-x)} &= \lim_{x\to 0}\frac{\ln(1-x^2)}{\ln(1+x)\ln(1-x)}\\ &= \lim_{x\to 0}\frac{-2x/(1-x^2)}{\ln(1-x)/(1+x)-\ln(1+x)/(1-x)}\\ &= \lim_{x\to 0}\frac{-2x}{(1-x)\ln(1-x)-(1+x)\ln(1+x)}\\ &= \lim_{x\to 0}\frac{-2}{-1-\ln(1-x)-1-\ln(1+x)}\\ &= \frac{-2}{-2}\\ &= 1. \end{align*}

share|cite|improve this answer
Nothing, apparently :S I gave up too quickly it seems. – stariz77 Apr 16 '12 at 1:54

Starting with Jim Conant's expression $$\frac{\ln(1-x^2)}{\ln(1-x)\ln(1+x)}$$ you could divide the numerator and the denominator by $x^2$, then use the fact that $$\lim_{x\to0} \frac{\ln(1+x)}{x} = 1.$$

share|cite|improve this answer

One approach is to use Taylor series. After combining fractions, you get $$\frac{\ln(1-x^2)}{\ln(1-x)\ln(1+x)}$$ Plugging in the Taylor series you get $$ \frac{-x^2-(1/2)x^4-\cdots}{(x-(1/2)x^2+\cdots)(-x-(1/2)x^2-\cdots)}=\frac{-1-(1/2)x^2+\cdots}{(1-(1/2)x+\cdots)(-1-(1/2)x+\cdots)} $$ This limits to $-1/-1=1$.

share|cite|improve this answer

Since $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$, we have $\log(1-x)=-x-\frac{x^2}{2}+O(x^3)$. Therefore, $$ \begin{align} \frac{1}{\log(1+x)}+\frac{1}{\log(1-x)} &=\frac{1}{x(1-\frac{x}{2}+O(x^2))}+\frac{1}{-x(1+\frac{x}{2}+O(x^2))}\\ &=\frac{1+\frac{x}{2}+O(x^2)}{x}-\frac{1-\frac{x}{2}+O(x^2)}{x}\tag{$\ast$}\\ &=\frac{x+O(x^2)}{x}\\ \end{align} $$ where $(\ast)$ is because $\frac{1}{1+x}=1-x+O(x^2)$.

Therefore, $$ \lim_{x\to0}\frac{1}{\log(1+x)}+\frac{1}{\log(1-x)}=1 $$

share|cite|improve this answer
$\lim_{x\to0}\frac{1}{\log(1+x)}+\frac{1}{\log(1-x)} $ the question is for + not -. – Mathlover Apr 16 '12 at 6:33
@Mathlover: The first '+' was a typo, but the rest was okay. I moved the '-' from the place which inspired the typo to the denominator. Since $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$, $\log(1-x)=-x-\frac{x^2}{2}+O(x^3)=-x(1+\frac{x^2}{2}+O(x^3))$. – robjohn Apr 16 '12 at 7:03
That's OK. I understood your solution. Really nice. I think ,in your first line , you have another typo too, it should be $\log(1-x)=-(x+\frac{x^2}{2}+O(x^3)$). Thanks – Mathlover Apr 16 '12 at 7:18
@Mathlover: Wow, for such a short answer, I certainly had a lot of typos. Thanks. – robjohn Apr 16 '12 at 11:31
You are welcome. +1 is for your different approach – Mathlover Apr 16 '12 at 11:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.