# Limits of Indeterminate Quotient

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!

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.

• Thanks for the clear clarification! I didn't think of applying L'Hopital Rule again.. – KillerKidz Mar 10 '14 at 17:11
• You are welcome. Simplest example $\lim_{x\to\infty} \frac{x^2}{e^x}$. Two L'H. Replace $x^2$ by $x^{17}$. Then $17$ L'H, though the pattern is obvious so one can just wave hands. – André Nicolas Mar 10 '14 at 17:15

Since we have

$$\ln(e^x-1)\sim_0\ln(1+x-1)=\ln x$$ then the desired limit is obviously $1$.

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}