Limits of Indeterminate Quotient: $\lim\limits_{x \to 0^+} \frac{\ln(e^x - 1)}{\ln(x)}$ and $\lim\limits_{x \to -1}(\frac1{x+1} - \frac3{x^3+1})$ I was preparing for my exam and found myself struggling with finding limits of indeterminate quotient.
$$\lim\limits_{x \to 0^+} \dfrac{\ln(e^x - 1)}{\ln(x)}$$
I have tried using L'Hopital's Rule to reduce it to:
$\lim\limits_{x \to 0^+} \dfrac{xe^x}{e^x-1}$
but still does not solve the problem.
Another problem that I've faced:
$$\lim\limits_{x \to -1}(\frac{1}{x+1} - \frac{3}{x^3+1})$$
I have tried to combine it into 1 term:
$\lim\limits_{x \to -1}(\dfrac{x^3-3x-2}{x^4+x^3+x+1})$
and applied L'Hopital's Rule but still got an Indeterminate Quotient.
Any advice on the 2 above qns is really much appreciated!
 A: For the first question, do it again! To simplify life, note that the $e^x$ on top approaches $1$, so you need not worry about it. Thus we want $\lim_{x\to 0^+}\frac{x}{e^x-1}$. Routine L'Hospital. 
For the second question, again remember that L'Hospital's Rule may need to be applied more than once. 
Remark: In the first question, we could have applied L'Hospital's Rule directly to $\frac{xe^x}{e^x-1}$. We mentioned the simplification because in more complicated situations, noticing this sort of thing can make computations much easier. 
A: Since we have
$$\ln(e^x-1)\sim_0\ln(1+x-1)=\ln x$$
then the desired limit is obviously $1$.
A: For the second limit you can also apply algebra.
$\displaystyle 
\begin{align}
\lim_{x \to -1}\left(\frac{1}{x+1} - \frac{3}{x^3+1}\right) &= \lim_{x \to -1}\left(\frac{1}{x+1} - \frac{3}{(x+1)(x^2-x+1)}\right) \\
 &= \lim_{x \to -1}\left(\frac{1}{x+1} - \left(\frac{1}{x+1}+\frac{2-x}{x^2-x+1}\right)\right) \\ 
 &= \lim_{x \to -1}\left(\frac{x-2}{x^2-x+1}\right)
\end{align}$
