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.
 A: We use:
i) $\lim\limits_{t \rightarrow 0} \frac{\sin t}{t}=1$,
ii) $\cos t - 1 = -2\sin^2\frac{t}{2}$.
Then
$\lim\limits_{x \rightarrow 0} \dfrac{\cos(\sin^2 x)-1}{\sin x}= \lim\limits_{x \rightarrow 0} \dfrac{-2\sin^2(\frac{\sin^2 x}{2})}{\sin x} = -\dfrac{1}{2}\lim\limits_{x \rightarrow 0} \dfrac{\sin^2(\frac{\sin^2 x}{2})}{(\frac{\sin^2 x}{2})^2}\sin^3 x = -\dfrac{1}{2}.(1)^2.0^3=0.$
A: Using
$$ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{2\sin^2(\frac {x}{2})}{x^2}=\frac12$$
one has
$$\lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x}=-\lim_{x \to 0} \frac{1-\cos\left(\sin^2x\right)}{\sin x}=-\lim_{x \to 0} \frac{1-\cos\left(\sin^2x\right)}{\sin^2x}\frac{\sin^2x}{\sin x}=0.$$
A: $\frac{\cos(\sin^2x)-1}{\sin x}=\frac{\cos(\sin^2x)-1}{\sin^4x}\cdot\sin^3x\stackrel{x\to 0}{\longrightarrow}\left(-\frac12\right)\cdot 0^3=0$
A: 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
$$
\lim_{x\to 0} \frac{\cos(\sin^2 x) - 1}{\sin x} = 0 .
$$
A: Using derivatives:
$$\begin{align*}
\lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x} &= \lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin^2(x)} \cdot \lim_{x\to0} \frac{\sin(x)}x \cdot \lim_{x\to0} x \\[1ex]
&= \frac{d\cos(y)}{dy}\bigg|_{y=0} \cdot \frac{d\sin(x)}{dx}\bigg|_{x=0} \cdot 0 & y=\sin^2(x) \\[1ex]
&= -\sin(0) \cdot \cos(0) \cdot 0 \\[1ex]
&= \boxed{0}
\end{align*}$$
