Evaluating $\lim\limits_{x \to 0} \frac{\cos\left(\sin^2 x\right)-1}{\sin x}$ without l'Hôpital I'm stuck trying to evaluate this limit without l'Hôpital's rule:
$$
 \lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x}
$$
Could anyone give me a hint on what I need to do? I tried many trig identities but I can't seem to find one that doesn't give me an indeterminate form.

My attempt, using asymptotic estimates.  As $u  \to 0$,
$$
\cos u = 1 -\frac12 u^2 + O(u^4) .
$$
As $x \to 0$, also $\sin x \to 0$, so
\begin{align}
\sin x &= x + O(x^3)
\\
\frac{1}{\sin x} &= \frac{1}{x}+O(x)
\\
\sin^2 x &= x^2 + O(x^4)
\\
\cos(\sin^2 x) &= 1 -\frac12 \sin^2 x + O(\sin^4 x) 
= 1 -\frac12 (x^2 + O(x^4)) + O(x^4) = 1 -\frac12 x^2 + O(x^4)
\\
\cos(\sin^2 x) - 1 &= -\frac12 x^2 + O(x^4)
\\
\frac{\cos(\sin^2 x) - 1}{\sin x} &=
\big(\frac{1}{x}+O(x)\big)\;\big(-\frac12 x^2 + O(x^4)\big)
= -\frac12 x + O(x^3)
\end{align}
so
$$
\lim_{x\to 0} \frac{\cos(\sin^2 x) - 1}{\sin x} = 0 .
$$
