# Limit of hyperbolic and trigonometric functions

$$\lim_{x\to0}\frac{\sinh x-\sin x}{x-\sin^2x}$$

As initially it's in 0/0 form, I applied L'Hôpital's rule.

$$\lim_{x\to0}\frac{\cosh x-\cos x}{1-\sin2x}$$

Now if I simply substitute $0$, then I get $0/1$ which is $0$. So is the answer $0$?

My book says answer is $1/3$.

it is $$\lim_{x \to 0}\frac{\cosh(x)-\cos(x)}{1-2\sin(x)\cos(x)}=0$$