What is the value of this improper integral? $\lim_{x\rightarrow 0 } \dfrac{1}{x} \int_{x}^{2x} e^{-t^2}\,\mathrm dt$ 
$$\lim_{x\rightarrow 0 } \dfrac{1}{x} \int_{x}^{2x} e^{-t^2}\,\mathrm dt$$

I don't have any idea to solve this integral.
 A: This is not an improper integral ($x\to 0$)
Now denote $f(x)=\int_0^xe^{-t^2}dt$
Keeping in mind $f(0)=0$ and $f'(x)=e^{-x^2}$, we're looking for:
$$\begin{align}\lim_{x\to 0}{f(2x)-f(x)\over x}&=2f'(0)-f'(0)\\&=1\end{align}$$
A: If you don't want to use L'Hôpital's rule, you can rewrite the limit as
$$
\lim_{h\rightarrow 0 } \frac{\int_{h}^{2h} e^{-t^2} dt-\int_{0}^{2\cdot0} e^{-t^2} dt}{h}
$$
and recognize the derivative of
$$
f(x):=\int_{x}^{2x} e^{-t^2} dt
$$
at $x=0$, which is (by the chain rule and the Fundamental Theorem of Calculus)
$$
f'(0)=2e^{0}-e^0=1
$$
A: In a neighbourhood of the origin:
$$ e^{-t^2} = 1-t^2 + o(t^3) \tag{1}$$
hence:
$$ \int_{x}^{2x}e^{-t^2}\,dt = x-\frac{7}{3}x^3+o(x^4) \tag{2}$$
and:
$$\frac{1}{x}\int_{x}^{2x}e^{-t^2}\,dt = \color{red}{1}-\frac{7}{3}x^2+o(x^3).\tag{3}$$
A: Perhaps, an easy way is using l'Hopital's rule:
$$\lim_{x\to 0}\frac{\int_x^{2x}e^{-t^2}\,dt}{x}=\lim_{x\to 0} (2e^{-4x^2}-e^{-x^2})=1$$
How do I differentiate the numerator?
You use the rule 

$$g(x)=\int_{a(x)}^{b(x)}f(t)\,dt\implies\\\implies g'(x)=b'(x)f(b(x))-a'(x)f(a(x))$$

which is valid whenever $f$ is continuous on a suitably big interval (say, $\mathbb R$) and $a,b$ are differentiable.
Indeed, $\int_{a(x)}^{b(x)}f(t)\,dt=F(b(x))-F(a(x))$ for a differentiable function such that $F'(x)=f(x)$. And you get the formula above using the chain rule.
A: Hint: Write down the series $1-t^2+\cdots$ for $e^{-t^2}$, integrate term by term, and divide by $x$.
Remark: If your question was about finding an antiderivative of $e^{-t^2}$ as a first step in solving the problem, that will not work. For $e^{-t^2}$ does not have an antiderivative that can be expressed in terms of elementary functions.
A: Solution based in squeezing: for $x>0$,
$$e^{-4x^2} = \frac1x (2x-x)e^{-(2x)^2}\le\dfrac{1}{x}\int_{x}^{2x}e^{-t^2}\, dt\le\frac1x(2x-x) e^{-x^2} = e^{-x^2}.$$
The case $x<0$ is similar.
A: More generally,
for small $x$,
if $f$ is differentiable
at zero,
$f(x)
\approx f(0)+xf'(0)
$
so
$\begin{array}\\
\int_{ax}^{bx} f(t) dt
&\approx \int_{ax}^{bx} (f(0)+tf'(0)) dt\\
&= (tf(0) + t^2f'(0)/2)\mid_{ax}^{bx}\\
&= (x(b-a)f(0) + x^2(b^2-a^2)f'(0)/2)\\
\text{or}\\
\frac1{x}\int_{ax}^{bx} f(t) dt
&\approx (b-a)f(0) + x(b^2-a^2)f'(0)/2)\\
&\to (b-a)f(0)
\qquad\text{as } x \to 0\\
\end{array}
$
If
$f(x) = e^{-x^2}$,
$a=1, b=2$,
then
$f(0) = 1$
so the limit is
$(2-1)1
= 1
$.
