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

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


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