Is it possible to apply L'Hopital's rule to this one ;



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

  • 2
    $\begingroup$ For the second, it is sadly all too possible, I have seen similar things done many times. But it gives the wrong answer. $\endgroup$ – 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.

  • $\begingroup$ 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 ? $\endgroup$ – optimal control Nov 6 '14 at 1:29
  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$ – Edward Jiang Nov 6 '14 at 1:29
  • $\begingroup$ @optimalcontrol Not entirely correct. You need to check the side limits on the second one. $\endgroup$ – 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 $$ \lim_{x\to 0} \frac{e^x}{1-e^x} = \mbox{does not exist}$$


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



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.