Calculate $\lim_{x\to 0} {1\over x} \int_0^x \cos(t^2)\,dt$

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 $x$.

Define $F(x) = \int_0^x\cos(t^2)\,dt$. By the fundamental theorem of calculus, $F'(x) = \cos(x^2)$ and so $$ \lim_{x\to 0}\frac 1x\int_0^x\cos(t^2)\,dt = \lim_{x\to 0}\frac{F(x)-F(0)}{x-0} = F'(0) =1. $$

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 $1$).