For a function defined by parts study continuity, and differentiability at two points

If you want to evaluate the limit:

$$\displaystyle\lim_{x\to 0}F(x)=\lim_{x\to 0}\int_{x}^{2x}\sin(t^2)dt$$

you can observe that $\forall x>0$ (the case $x<0$ is the same), $f(t)=\sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $\xi_{x}\in (x,2x)$ such that

$$\int_{x}^{2x}\sin(t^2)dt=\sin(\xi_{x}^2)(2x-x)\implies F(x)=\sin(\xi_{x}^2)x$$

Now $\xi_{x}\to 0$ for $x\to 0^{+}$ so:

$$\lim_{x\to 0^{+}}F(x)=\lim_{x\to 0}\sin(\xi_{x}^2)x=[\sin(0)\cdot 0]=0$$

Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:

$$\lim_{x\to 0^{+}}\frac{F(x)-F(0)}{x-0}=\lim_{x\to 0}\frac{\sin(\xi_{x}^2)x}{x}=\lim_{x\to 0}\sin(\xi_{x}^2)=0$$

For $x_0=\sqrt{\frac{\pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.

$F'(x)=2\sin(4x^2)-\sin(x^2)\implies F'\left(\sqrt{\frac{\pi}{2}}\right)=2\sin(4\cdot\frac{\pi}{2})-\sin(4\cdot\frac{\pi}{2})=-1$