Calculate $\lim_{x\to 0} {1\over x} \int_0^x \cos(t^2)\,dt$ I have to calculate $$\lim_{x\to 0} {1\over x} \int_0^x \cos(t^2)\,dt$$
My intuition is that the answer is 1 because as $x$ becomes very small, $x^2$ also becomes very small and I am tempted to write $$\lim_{x\to 0} \int_0^x \cos(t^2)\,dt=\lim_{x\to 0} \int_0^x \cos(t)\,dt$$
And then, because $(\sin x)'=\cos x$,
we have $$\lim_{x\to 0} {1\over x} \int_0^x \cos(t^2) \, dt = \lim_{x\to 0} {1\over x} \int_0^x \cos(t) \, dt=\lim_{x\to 0} {1\over x} (\sin x -\sin 0)=\lim_{x\to 0} {\sin x\over x}= 1$$
(well known limit solved with L'Hôpital's rule)
But I am pretty sure this is not rigorous and I am doing something I am not 'allowed to', especially the first equality I wrote.
 A: Or go through by Integral Mean Value:
\begin{align*}
\dfrac{1}{x}\int_{0}^{x}\cos(t^{2})dt=\cos(\eta_{x}^{2})\rightarrow\cos 0=1,
\end{align*}
where $\eta_{x}$ is in between $0$ and $x$.
A: Define $F(x) = \int_0^x\cos(t^2)\,dt$. By the fundamental theorem of calculus, $F'(x) = \cos(x^2)$ and so
$$
\lim_{x\to 0}\frac 1x\int_0^x\cos(t^2)\,dt = \lim_{x\to 0}\frac{F(x)-F(0)}{x-0} = F'(0) =1. 
$$
A: The fundamental theorem of calculus tells you that $$ \frac d {dx} \int_0^x \cos(t^2)\, dt = \cos(x^2).$$
The usual definition of derivative tells you that $$ \lim_{x\to0} \frac{\int_0^x \cos(t^2) \, dt - \int_0^0 \cos(t^2)\, dt}{x-0} = \left.\frac d {dx} \int_0^x \cos(t^2) \, dt \right|_{x\,=\,0}.$$
So you get $\cos(0^2)$ (which is $1$).
A: I'd say you're doing too much work; by L'Hopital's rule and the fundamental theorem of calculus,
$$\lim_{x\to0}\frac{\int_0^x\cos(t^2)\,\mathrm dt}x=\lim_{x\to0}\frac{\cos(x^2)}1=1$$
A: Of course, this answer is overkill, but it does illustrate the power of Lebesgue's Density theorem, to take care of exercises like this one. Using that theorem and the fact that cos is even
\begin{align}\dfrac{1}{x}\int_{0}^{x}\cos(t^{2})dt=\dfrac{1}{2x}\int_{-x}^{x}\cos(t^{2})dt\to \cos 0=1.\end{align}
