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?
 A: 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*}
A: 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
$$\lim_{x\to0} \frac{\ln(1+x)}{x} = 1.$$
A: 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 $-1/-1=1$.
A: 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,
$$
\lim_{x\to0}\frac{1}{\log(1+x)}+\frac{1}{\log(1-x)}=1
$$
