Find $$\lim_{x\to 0} \frac{\sinh x -\sin x}{\cosh x - \cos x}$$
I used Hopital and couldn't find the answer which is $0$.
Find $$\lim_{x\to 0} \frac{\sinh x -\sin x}{\cosh x - \cos x}$$
I used Hopital and couldn't find the answer which is $0$.
Using Taylor series, $\sinh x-\sin x=\frac{x^3}{3}+o(x^3)$ while $\cosh x-\cos x=x^2+o(x^2)$, so the fraction is $\frac{x}{3}+o(x)$. This proves the limit is $0$, whether we take $x\to 0^+$ or $x\to 0^-$.
Hint. Apply L'Hopital's rule TWO times: $$\lim_{x\to 0} \frac{\sinh x -\sin x}{\cosh x - \cos x}\stackrel{H}{=}\lim_{x\to 0} \frac{\cosh x -\cos x}{\sinh x + \sin x}\stackrel{H}{=}\lim_{x\to0} \frac{\sinh x +\sin x}{\cosh x + \cos x}=?$$