Given $F(x)=\int\limits_x^{x^2}\frac{\sin t}{t}dt$, find $\lim_{x\rightarrow 0}F(x)$ and $\lim_{x\rightarrow 0}F'(x)$ 
Given $F(x)=\int\limits_x^{x^2}\frac{\sin t}{t}dt$, find $\lim_{x\rightarrow 0}F(x)$ and $\lim_{x\rightarrow 0}F'(x)$
Assume the options are ${-1, 0, 1}$

Intuitively I'm pretty sure the answer is $\lim_{x\rightarrow 0}F(x)=0$ , because $\frac{\sin t}{t}$ approaches 1 and we look at a smaller segment as $x$ approaches $0$, so the area under the graph would approach $0$.
Also, I'm pretty sure $\lim_{x\rightarrow 0}F'(x)=-1$, because $x^2 < x$ for a small enough $x$ so we would get minus the value of $\frac{\sin 0}{0}$.
I'm having trouble formalizing this, would appreciate help with finding more precise arguments, and also of course let me know if I'm wrong.
 A: Here is a solution for the first limit using only the the fact that $\lim_{t\rightarrow0}\frac{\sin{t}}{t}=1$. 
For some $\delta>0$, $\Big|\frac{\sin t}{t}\Big|\leq \frac{3}{2}$ whenever $|t|<\delta$.  So for $|x|<\min(\delta,1)$, $x^2\leq|x|$ and
$$
\left|\int^{x^2}_x\frac{\sin t}{t}\,dt\right|\leq \frac{3}{2}|x-x^2|\leq 3|x|
$$
Letting $x\rightarrow0$, you obtained that the first limit you are looking for is indeed $0$. The second is straight-forward differentiation exercise as pointed out in an early solution.
A: I will start with the second limit. You can easily evaluate the derivative of $F$ as follows: let $g(t):= \frac{\sin t}{t}$, and note that $F(x)=G(x^2)-G(x)$. Then, also computing the derivative of the composite function, you obtain:
$$ F'(x)=2x g(x^2) - g(x) = 2x \cdot \frac{\sin x^2}{x^2} - \frac{\sin x}{x} $$
Now, recall that:
$$ \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$ 
This gives you:
$$ \lim_{x \rightarrow 0} F'(x) = 2 \cdot 0 \cdot 1 - 1 = -1 $$
For the other limit, consider the MacLaurin series of the integrand, and integrate term by term. It is not difficult to see that the result of the limit is $0$.
EDIT: we have the Taylor series:
$$ \frac{\sin t}{t} = \sum_{k=0}^{\infty} \frac{(-1)^n}{(2n+1)!} t^{2n} $$
Now integrate term by term (it is known that this is possible in a case like this one):
$$ F(x) = \int_{x}^{x^2} \frac{\sin t}{t} dt = \sum_{k=0}^{\infty} \int_{x}^{x^2} \frac{(-1)^n}{(2n+1)!} t^{2n} dt = \sum_{k=0}^{\infty} \frac{(-1)^n}{(2n+1)! (2n+1)} x^{2n+1} (x^{2n+1} - 1) $$
Then, you can see that, as $x \rightarrow 0$, $F(x) \rightarrow 0$.
EDIT: another solution. Note that $g(t)=\frac{\sin t}{t}$ is bounded. To see this, recall that $|\sin t | \leq |t|$, which implies that $|g(t)| \leq 1$, and thus the function is indeed bounded. Then, we can write:
$$ \left |\int_{x}^{x^2} \frac{\sin t}{t} dt \right | \leq |x^2 - x| \cdot 1 = |x^2 - x| $$
When $x \rightarrow 0$, the RHS goes to $0$, and thus:
$$ \lim_{x \rightarrow 0} F(x) = 0 $$
