evaluating $\lim_{x \rightarrow 0} \frac{\int_{0}^{2\sin x} \cos(t^2) dt}{2x}$ I'm having a hard time evaluating the following limit
$$\lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}$$
I'm not even really sure how to approach it since I'm not used to seeing another function in the integral.
My initial thoughts were to use l'hopitals rule since $\lim_{x \rightarrow0} 2\sin x=0$, so the numerator is equal to 0, but I wouldnt be sure how to take the derivative of the numerator. 
Any pointers would be appreciated
 A: Without Liebnitz rule, you can use first mean value theorem for integrals, then there exist $c$ such that
$$\lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}=\lim_{x \rightarrow 0} \frac{2 \sin x \cos(c^2) }{2x}=\lim_{x \rightarrow 0} \frac{2 \sin x }{2x}\lim_{x \rightarrow 0} \cos(c^2)=\lim_{x \rightarrow 0} \cos(c^2)=1$$
because with sandwich $0\leq c\leq2\sin x$ so $c\to0$ as $x\to0$.
A: The denominator $2x$ can be replaced by $2\sin x$ via the limit $\lim_{x\to 0}(\sin x) /x=1$ and then putting $u=2\sin x$ the limit is easily seen to be $$\lim_{u\to 0} \frac{1}{u}\int_{0}^{u}\cos(t^2)\,dt=\cos(0^2)=1$$ via Fundamental Theorem of Calculus. 
A: We will use that $$f(x)=\int_0^{a(x)}g(t)dt\implies f'(x)=g(a(x))a'(x).$$ In our case:
$$\frac{d}{dx} \int_{0}^{2 \sin x} \cos(t^2) dt=\cos((2\sin x)^2)\cdot \dfrac{d(2\sin x)}{dx}.$$
So, we have
$$\frac{d}{dx} \int_{0}^{2 \sin x} \cos(t^2) dt=2\cos (x)\cos(4\sin^2 x).$$
Thus
$$\lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}=\lim_{x \rightarrow 0}\dfrac{2\cos (x)\cos(4\sin^2 x)}{2} =1.$$
A: Alternatively, use Taylor expansion:
$$\cos(t^2)=1-\frac{t^4}{2}+\frac{t^8}{24}+O(t^{9});\\
\int_{0}^{2\sin x} \cos(t^2)dt=\left(t-\frac{t^5}{10}+O(t^6)\right)\bigg{|}_0^{2\sin x}=2\sin x-\frac{(2\sin x)^5}{10}+O((2\sin x)^6);\\
\lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}=\lim_{x \rightarrow 0}\frac{2\sin x+O(\sin^5x)}{2x}=\lim_{x \rightarrow 0}\frac{2\sin x}{2x}\cdot (1+O(\sin^4x))=1.$$
