Hi

Evaluate the following limit :

$\displaystyle \lim_{x\to0} \frac{sinh(x) - sin(x)}{sin^3(x)}$

I used L`Hospital`s, But I stucked.

I tried to use the sandwich theorem:

Clearly $\displaystyle |\frac{sinh(x) - sin(x)}{sin^3(x)}| \leq -(sinh(x)-sin(x))$

Since the minumum value for $\displaystyle sin^3(x)$ is $\displaystyle -1$.

since I make the denominator smaller then the whole fraction is bigger.

and since $\displaystyle -\lim_{x\to0} (sinh(x)-sin(x)) = 0$

then ,By using the sandwich theorem, the desired limit $\displaystyle =0$

Is this right?

Do you have another way to make the limit simpler ?