Limit of a function ("order of magnitude")

Question

Find $f(x)$ (in terms of exponentials) such that $$\lim_{x \to 0}\frac{e^{-\cosh\frac{1}{x}}}{f(x)}=A$$ where $A\in\mathbb{R}, A\neq0$

I have tried calculating Maclaurin series but I get zeroes everywhere, like $e^{-cosh\frac{1}{x}}$ is not analytic. I don't know what else to try!

Answer 1

Note that

$$\cosh\frac{1}{x}=\frac{e^{\frac1x}+e^{-\frac1x}}{2}=\frac12e^{\left|\frac1x\right|}+\frac12e^{-\left|\frac1x\right|} $$

thus

$$e^{-\cosh\frac{1}{x}}= e^{\frac12e^{\left|\frac1x\right|}+\frac12e^{-\left|\frac1x\right|}}= e^{\frac12e^{\left|\frac1x\right|}} e^{ \frac12e^{-\left|\frac1x\right|}}\sim e^{\frac12e^{\left|\frac1x\right|}}$$

and

$$f(x)= e^{-\frac12e^{\left|\frac1x\right|}}\implies \lim_{x \to 0}\frac{e^{-\cosh\frac{1}{x}}}{f(x)}=1$$