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

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?

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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*}

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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 $$\lim_{x\to0} \frac{\ln(1+x)}{x} = 1.$$

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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 $-1/-1=1$.

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

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