1
$\begingroup$

I need to show $$\lim_{x\rightarrow 0^+}\frac{1+\cos x}{e^x-1}=\infty$$

I know that, say, if you let $f(x) = 1 + \cos x$ and $g(x) = \dfrac{1}{e^x-1}$, and then multiply the limits of $f(x)$ and $g(x)$, you get $\frac{2}{0}$. I can't figure out how to make it work for l'Hopital's rule however, i.e. how to rewrite it so that it is in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

I also tried multiplying $h(x)$ by the conjugate of $f(x)$, but I don't think this is fruitful. Any hints appreciated.

$\endgroup$
4
$\begingroup$

I don't know why you like that form. If you insist, $$ \lim_{x\to 0^+} \frac{(\cos x +1)}{e^x-1} $$ can be rewritten to $$\lim_{x\to 0^+} \frac{2+(\cos x-1)}{e^x-1}$$ Then you can use L'Hopital rule with the right part. It seems wired.

$\endgroup$
1
$\begingroup$

Recall that

$$e^x \sim 1 + x + \text{(high order terms)},$$

for $x \to 0^+$.

Then $e^x - 1 \sim x$, and you can solve:

$$\lim_{x\rightarrow0+}\frac{(1+\cos x)}{x} = \ldots$$

$\endgroup$
  • $\begingroup$ Are we justified in lopping off the higher order terms because we are near the origin? $\endgroup$ – AndyDufresne Nov 3 '17 at 10:22
  • $\begingroup$ To be more precise, the same job can be done for the numerator. In this case $1+ \cos x \sim 1 + 1 - \frac{x^2}{2} + \text{hot}$. Then, the whole thing reduces to: $$\frac{2 - \frac{x^2}{2} + \text{hot}}{x + \text{hot}}.$$ Then, yes, you are allowed to do this. Notice that $\text{hot} \to 0$ as $x \to 0$. $\endgroup$ – the_candyman Nov 3 '17 at 10:25

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.