How do I evaluate this limit $\lim_{x\to \infty}\frac{1}{x}\int_x^{2x}e^{-t^2}dt$ without using the mean value theorem for integrals? How do I evaluate this limit $$\lim_{x\to \infty}\frac{1}{x}\int_x^{2x}e^{-t^2}dt$$
without using the mean value theorem for integrals? Is taking the derivative of this with respect to $x$ going to help me at all? Apart from that, I was thinking Riemann sums. Other than that I'm stuck, and just looking for a quick nudge in the right direction.
 A: You have 
$$
0\leq \frac{1}{x}\int_x^{2x}e^{-t^2}\,dt\leq\frac{1}{x}\int_x^{2x}e^{-x^2}\,dt=e^{-x^2}.
$$
A: The integrand is decreasing, so by the trivial/M-L bound/positivity of the integral/whatever, we have
$$ 0 \leq \frac{1}{x} \int_x^{2x} e^{-t^2} \, dt \leq \frac{1}{x} (2x-x) e^{-x^2} = e^{-x^2} , $$
which is enough to show the limit is $0$.
A: For $ x$ great enough, put
$$u=\frac{t}{x}$$
it becomes
$$0\le \lim_{x\to +\infty}\int_1^2e^{-u^2x^2}du\le \lim_{x\to+\infty} e^{-x^2}$$
because
$$1\le u \le 2$$ and
$$-x^2u^2\le -x^2$$
A: Hospital;
$\lim_{x \rightarrow \infty} \frac{2e^{-4x^2}-e^{-x^2}}{1}=0$.
Or:
$(1/x)\displaystyle{\int_{x}^{2x}}e^{-t^2}dt <(1/x)\int_{x}^{2x}e^{-t}dt$
$=(1/x)(-e^{-2x}+e^{-x}).$
A: $$\frac{1}{x}\int_x^{2x}e^{-t^2}dt$$
$$\frac{1}{x}\int_0^{2x}e^{-t^2}dt-\frac{1}{x}\int_0^xe^{-t^2}dt.$$ Let $F(x)=\int_0^{2x}e^{-t^2}dt$,  $G(x)=\int_0^xe^{-t^2}dt.$ Then $$F'(x)=2e^{-4x^2}, F'0)=2$$ By the definition of derivative, $$F'(0)=\lim_{x \to 0}\frac{F(x)-F(0)}{x-0}=\lim_{x \to 0}\frac{F(x)}{x}$$
$$=\lim_{x \to 0}\frac{1}{x}\int_0^{2x}e^{-t^2}dt.$$ Thus
$$\lim_{x \to 0}\frac{1}{x}\int_0^{2x}e^{-t^2}dt=2.$$ Similarily, $$G'(x)=e^{-x^2}, G'0)=1$$ By the definition of derivative, $$G'(0)=\lim_{x \to 0}\frac{G(x)-G(0)}{x-0}=\lim_{x \to 0}\frac{G(x)}{x}$$
$$=\lim_{x \to 0}\frac{1}{x}\int_0^{x}e^{-t^2}dt.$$ Thus
$$\lim_{x \to 0}\frac{1}{x}\int_0^xe^{-t^2}dt=1.$$
$$\text{Therefore, }\lim_{x \to 0}\frac{1}{x}\int_x^{2x}e^{-t^2}dt$$
$$=\lim_{x \to 0}\frac{1}{x}\int_0^{2x}e^{-t^2}dt$$
$$-\lim_{x \to 0}\frac{1}{x}\int_0^xe^{-t^2}dt$$
$$=2-1=1.$$
