Evaluate $ \lim\limits_{x \to 0} \frac{1}{\sin^3x}\int_0^x{\sin(t^2) } dt$ 
$$ \lim_{x \to 0} \frac{1}{\sin^3x}\int_0^x{\sin(t^2) dt}$$  

This is what I've tried: 
Let $F(x) = \displaystyle\int_0^x{\sin(t^2) dt}$, and let $f(x) = {\sin(t^2)}$.
Then $F'(x) = f(x) \Rightarrow F'(0) = f(0) = \sin 0 = 0$. 
Where do I go from here?
Thanks for help.
 A: There are many ways. My first choice would be to multiply by $\frac{x^3}{x^3}$ to get 
$$\frac{x^3}{\sin^3 x}\frac{1}{x^3}\int_0^x \sin(t^2)\,dt,$$
and then expand $\sin(t^2)$ in a power series, and integrate term by term.
It is a standard fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$, and therefore $\lim_{x\to 0}\frac{x^3}{\sin^3 x}=1$. For the rest, the power series for $\sin(t^2)$ is $t^2-\frac{t^6}{3!}+\cdots$, and integration gives $\frac{x^3}{3}$ plus terms that go to $0$ much faster than $x^3$, 
A: Using L'Hospital's rule once, we have
$$
\lim_{x\rightarrow 0}\frac{\sin x^2}{3\sin^2x \cos x}.
$$
We examine the limit
$$
\lim_{x\rightarrow 0} \frac{\sin x^2}{\sin^2x} = \lim_{x\rightarrow 0} \frac{2x\cos x^2}{2\sin x \cos x} =\lim_{x\rightarrow 0} \frac{x\cos x}{\sin x} = 1
$$
using the fact that $\lim_{x\rightarrow 0} \frac{x}{\sin x} = 1$. Thus,
$$
\lim_{x\rightarrow 0}\frac{\sin x^2}{3\sin^2x \cos x} = \lim_{x\rightarrow 0} \frac{1}{3\cos x} = \frac{1}{3}.
$$
A: Use L'Hopital's rule. The derivative of the denominator is easy to find.
A: Let $F'(t)=\sin(t^2)$. Then we have
$$
\lim_{x\to 0}\frac{F(x)-F(0)}{\sin^3x}.
$$
Using L'Hospital rule and $\lim_{x\to 0}\sin x/x=1$ we have
$$
\lim_{x\to 0}\frac{F(x)-F(0)}{\sin^3x}=\lim_{x\to 0}\frac{F'(x)}{3\sin^2 x \cos x}
$$
$$
=\lim_{x\to 0}\frac{x^2\sin(x^2)}{3x^2\sin^2 x \cos x}=\frac 1 3.
$$
A: Hint: break it up.  There are some missing pieces for you to fill in.  
$$\lim_{x\to 0}\frac{1}{\sin^3 x} \int_0^x \sin(t^2)\,dt = \dots (\lim_{x \to 0}\frac{x}{\sin x})^3 \lim_{x\to 0}  \frac{\int_0^x \sin(t^2)\,dt}{x^3}.$$
Use L'Hopital's Rule on the left factor (inside the cube).  Use L'Hopital's Rule and the Fundamental Theorem of Calculus on the right factor:
$$\lim_{x\to 0}  \frac{\int_0^x \sin(t^2)\,dt}{x^3} = 
  \lim_{x\to 0}  \frac{\sin(x^2)}{3x^2}. $$
Substitute $t = x^2$:
$$ \lim_{x\to 0}  \frac{\sin(x^2)}{3x^2} = \lim_{t\to 0^+}  \frac{\sin(t)}{3t}. $$
Then apply L'Hopital's Rule again.
