# 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

$\lim_{x\rightarrow 0}\frac{e^{x}}{1-e^{x}}$

• For the second, it is sadly all too possible, I have seen similar things done many times. But it gives the wrong answer. – André Nicolas Nov 6 '14 at 1:21

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.

• Thanks for the helpful answer. So, the limit for the first one will be -1 and the limit for the second will be infinity, am I right ? – optimal control Nov 6 '14 at 1:29
• Yes, that is correct. – Edward Jiang Nov 6 '14 at 1:29
• @optimalcontrol Not entirely correct. You need to check the side limits on the second one. – Aaron Maroja Nov 6 '14 at 1:37

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 $$\lim_{x\to 0} \frac{e^x}{1-e^x} = \mbox{does not exist}$$

$$L=\lim_{x\to\infty}\frac{e^x}{1-e^x}\sim\frac{+\infty}{-\infty}\implies L=\lim_{x\to\infty}\frac{e^x}{-e^x}=-1$$

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