L'Hopital's rule for $\frac{e^{x}}{1-e^{x}}$ for$ x\to 0, \infty$ Is it possible to apply L'Hopital's rule to this one ;
$\lim_{x\rightarrow\infty}\frac{e^{x}}{1-e^{x}}$
and 
$\lim_{x\rightarrow 0}\frac{e^{x}}{1-e^{x}}$
 A: Yes, you can apply L'Hopital's rule to the first one, but not the second one. The reason is because the first one
$$\lim_{x\to \infty} \frac{e^x}{1-e^x}$$
is of the indeterminate form
$$-\frac{\infty}{\infty}$$
but the second one 
$$\lim_{x\to 0} \frac{e^x}{1-e^x}$$
is of the form
$$\frac{1}{0}$$
which is simply undefined. 
A: L'Hospital's Rule
Assuming that $f(x)$ and $g(x)$ are differentiable, $\frac{d}{dx}g(x)\neq 0$ and
$$ \lim_{x\to c} \frac{f(x)}{g(x)}= \frac{0}{0}\quad \mbox{or}\quad  \lim_{x\to c} \frac{f(x)}{g(x)}= \frac{\pm\infty}{\pm\infty} $$
Then, 
$$ \lim_{x\to c} \frac{f(x)}{g(x)}= \lim_{x\to c} \frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)}=L $$
Where $c$ and $L$ is any real number or $\pm\infty$.

The first limit meets all of the above conditions, so we have
$$ \lim_{x\to\infty} \frac{e^x}{1-e^x} = \lim_{x\to\infty} \frac{\frac{d}{dx}\left[e^x\right]}{\frac{d}{dx}\left[1-e^x\right]}= \lim_{x\to\infty} \frac{e^x}{0-e^x} =\lim_{x\to\infty} -\frac{e^x}{e^x} = -1$$
However, the second limit does not meet the above conditions so we cannot apply L'Hospital's rule. So now lets check the left and right sided limits, 
$$ \lim_{x\to 0^-} \frac{e^x}{1-e^x} = \infty$$
And
$$ \lim_{x\to 0^+} \frac{e^x}{1-e^x} = -\infty$$
Therefore
$$ \lim_{x\to 0} \frac{e^x}{1-e^x} = \mbox{does not exist}$$
A: 
$$L=\lim_{x\to\infty}\frac{e^x}{1-e^x}\sim\frac{+\infty}{-\infty}\implies L=\lim_{x\to\infty}\frac{e^x}{-e^x}=-1$$



$$L=\lim_{x\to0}\frac{e^x}{1-e^x}\sim\frac{1}{0}\sim\infty[\text{undefined}]$$

