How to calculate this limit involving an integral I missed a couple of classes so I'm having trouble doing this (and other similar) excercises of homework:
$\lim\limits_{x\to0^+}\frac{\displaystyle\sqrt{x}-\displaystyle\int_0^\sqrt{x} \mathrm{e}^{-t^2}\,\mathrm{d}t}{\displaystyle\sqrt{x^3}}$
I am supposed to solve this limit using De L'Hospital rule, however I don't know how to take the derivative of the numerator.
Using the fundamental theorem of calculus I would know how to derivate $\displaystyle\int_0^x \mathrm{e}^{-t^2}\,\mathrm{d}t$, but I don't know what to do when I have a function of $x$ as a limit of the integral, even though I guess it's related to the fundamental theorem.
Is there a general rule to calculate the first derivative of $\displaystyle\int_a^{f(x)} f(t)\,\mathrm{d}t$ ?
 A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#00f}{\large\lim_{x \to 0^{+}}%
{\displaystyle{\root{x} - \int_{0}^{\root{x}}\expo{-t^{2}}\,\dd t} \over \displaystyle{\root{x^{3}}}}}
=\lim_{x \to 0^{+}}%
{\displaystyle{x - \int_{0}^{x}\expo{-t^{2}}\,\dd t} \over \displaystyle{x^{3}}}
=\lim_{x \to 0^{+}}%
{\displaystyle{1 - \expo{-x^{2}}} \over \displaystyle{3x^{2}}}
\\[3mm]&=\lim_{x \to 0^{+}}%
{\displaystyle{2x\expo{-x^{2}}} \over \displaystyle{6x}}
={1 \over 3}\lim_{x \to 0^{+}}\expo{-x^{2}}
=\color{#00f}{\large{1 \over 3}}
\end{align}
A: You mean "calculate the derivative of $\int_a^{g(x)} f(t) dt$ with respect to $x$", in which case yes you can in this way:
Let $F(x) = \int_a^{x} f(t) dt$
Then $F'(x) = f(x)$  [by the fundamental theorem]
Thus $\frac{d}{dx}( F(g(x)) ) = (Fg)'(x) = F'(g(x)) g'(x) = f(g(x)) g'(x)$  [by chain rule]
It is good that you ask about the general question, but for this special case there is actually a faster way, because we can substitute $y=\sqrt{x}$, and $x \to 0^+$ is equivalent to $y \to 0^+$. Then the expression simplifies and we can apply the fundamental theorem directly.
A: If you know enough about series and integrals to justify swapping of order between $\sum$ and $\int$ (or if you want to get a sense of what the limit is before trying to prove its value in a different way), you can do it this way (not the best way, but an entertaining one):


*

*recall that for all $x\in\mathbb{R}$,  $e^x \stackrel{\rm{}def}{=}\sum_{n=0}^\infty \frac{x^n}{n!}$;

*write 
\begin{align*}
\sqrt{x} - \int_0^{\sqrt{x}}e^{-t^2}dt &= \int_0^{\sqrt{x}}\left( 1-e^{-t^2} \right)dt \\
&= \int_0^{\sqrt{x}}\left( 1-\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{n!} \right)dt \\
&= \int_0^{\sqrt{x}}\left( 1-1-\sum_{n=1}^\infty \frac{(-1)^n t^{2n}}{n!} \right)dt \\
&= -\int_0^{\sqrt{x}}\sum_{n=1}^\infty \frac{(-1)^n t^{2n}}{n!} dt \\
&= -\sum_{n=1}^\infty \frac{(-1)^n }{n!} \left( \int_0^{\sqrt{x}} t^{2n} dt\right)\tag{swap $(\dagger)$} \\
&= -\sum_{n=1}^\infty \frac{(-1)^n }{n!} \frac{x^{(2n+1)/2}}{2n+1} \\
&= \frac{x^{3/2}}{3} - \sum_{n=2}^\infty \frac{(-1)^n }{n!} \frac{x^{(2n+1)/2}}{2n+1}
\end{align*}
so that 
$$
\frac{\sqrt{x}-\int_0^\sqrt{x} {e}^{-t^2}dt}{\sqrt{x^3}} = \frac{1}{3} - \sum_{n=2}^\infty \frac{(-1)^n }{n!} \frac{x^{(2n+1)/2-3/2}}{2n+1} = \frac{1}{3} - \sum_{n=2}^\infty \frac{(-1)^n }{n!} \frac{x^{n-1}}{2n+1} = \frac{1}{3} + o(1)
$$
(where the very last equality, $(\ddagger)$, also needs a brief justification). Hence, and modulo arguing$(\dagger)$ and $(\ddagger)$, this shows the limit is $1/3$.

A: Other than Newton leibnitz theorem in differentiating the integral you have to use l' hospital's rule. i.e limit of f(x)/g(x) as x approaches to 0 is equal to limit of f'(x)/g'(x) as x approaches to 0
