Limit of a function with log Find the value of 
$$\lim_{t\to 0}\left(\frac{1}{\ln(1 + t)}+\frac{1}{\ln(1-t)}\right).$$
I tried L'Hospital's Rule but can't get it to work. I can't seem to find the right algebra tricks to apply before attempting L'Hospitals' Rule.
 A: Bring to a common denominator. Since $\ln(1-t)+\ln(1+t)=\ln(1-t^2)$ we want to find
$$\lim_{t\to 0}\frac{\ln(1-t^2)}{\ln(1+t)\ln(1-t)}.$$
Now apply L'Hospital's Rule. We end up wanting
$$\lim_{t\to 0}\frac{\frac{-2t}{1-t^2}}{\frac{\ln(1-t)}{1+t}-\frac{\ln(1+t)}{1-t}}.$$
Multiply top and bottom by $(1-t)(1+t)$. We want
$$\lim_{t\to 0} \frac{-2t}{(1-t)\ln(1-t)-(1+t)\ln(1+t)}.$$
Now apply L'Hospital's Rule again. The derivative of the top is the harmless $-2$. The derivative of the bottom is $-1-\ln(1-t)-1-\ln(1+t)$, which has limit $-2$. Now  we are finished, the limit is $1$. 
Remark: I would not use L'Hospital's Rule. The power series expansions of $\ln(1+t)$ and $\ln(1-t)$ work much more smoothly. We have $\ln(1+t)=t-\frac{t^2}{2}+O(t^3)$ and $\ln(1-t)=-t-\frac{t^2}{2}+O(t^3)$, and now basic algebra does it. 
A: $$\dfrac{1}{\ln(1+t)}+\dfrac{1}{\ln(1-t)}=\dfrac{\ln(1-t)+\ln(1+t)}{\ln(1+t)\ln(1-t)}$$
Implying that
$$\lim_{t\to0}\dfrac{\ln(1-t)+\ln(1+t)}{\ln(1+t)\ln(1-t)}=\lim_{t\to0}\dfrac{\ln(1-t^2)}{\ln(1+t)\ln(1-t)}=-\lim_{t\to0}\dfrac{\frac{2t}{1-t^2}}{\frac{1}{1-t}-\frac{1}{1+t}}=\\
-\lim_{t\to0}\dfrac{\frac{2t}{1-t^2}}{\frac{1+t-1+t}{1-t^2}}=-1$$
A: As pointed by André Nicolas, you could also simply use the Taylor expansion $$\log(1+x)\simeq \frac{x}{1}-\frac {x^2}{2}$$ So $$\frac{1}{\ln(1 + t)}+\frac{1}{\ln(1-t)}\simeq \frac{t^2}{(t-\frac{t^2}{2})(t+\frac{t^2}{2})}=\frac{t^2}{t^2-\frac{t^4}{4}}=\frac{1}{1-\frac{t^2}{4}}$$
