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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.

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2 Answers 2

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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.

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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:

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

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  • $\begingroup$ Are we justified in lopping off the higher order terms because we are near the origin? $\endgroup$
    – AndyDufresne
    Nov 3, 2017 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, 2017 at 10:25

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