Limit by L'hospital's rule I have to prove that: 
 $$\lim \limits_{x \to \infty} \frac{{\int_x^{\infty} \exp(-t^2/2)dt}}{\exp(-x^2/2) 
(1/x)}=1$$   
Should I use L'hospital rule, if yes what are the derivatives?  
 A: First note that
$$ 
\int e^{-cx^2}dx=\sqrt{\frac{\pi}{4c}}\mathrm{erf}(\sqrt{c}x)
$$
So now we have
$$
\lim_{x \to \infty} \left[\frac{\int_x^{\infty} e^{-\frac{t^2}{2}}dt}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\lim_{x \to \infty} \left[\frac{\lim \limits_{b\to\infty} \int_x^b e^{-\frac{t^2}{2}}dt}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\lim_{x \to \infty} \left[\frac{\sqrt{\frac{\pi}{2}}\lim \limits_{b\to\infty}\left[\mathrm{erf}\left(\frac{b}{\sqrt{2}}\right)-\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)\right]}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]
$$
$$
=\lim_{x \to \infty} \left[\frac{\sqrt{\frac{\pi}{2}}\left[1-\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)\right]}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\lim_{x \to \infty} \left[\frac{\sqrt{\frac{\pi}{2}}\mathrm{erfc}\left(\frac{x}{\sqrt{2}}\right)}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\mathrm{erfc}\left(\frac{x}{\sqrt{2}}\right)}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]
$$
$$
=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\frac{d}{dx}\mathrm{erfc}\left(\frac{x}{\sqrt{2}}\right)}{\frac{d}{dx}\left[\frac{1}{x}e^{-\frac{x^2}{2}}\right]}\right]=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{-\sqrt{\frac{2}{\pi}}e^{-\frac{x^2}{2}}}{-\frac{e^{-\frac{x^2}{2}}(x^2+1)}{x^2}}\right]=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\sqrt{\frac{2}{\pi}}}{\frac{(x^2+1)}{x^2}}\right]
$$
$$
=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\sqrt{\frac{2}{\pi}}}{1+\frac{1}{x^2}}\right]=\sqrt{\frac{\pi}{2}}\sqrt{\frac{2}{\pi}}=1
$$
A: Since 
$$\mathop {\lim }\limits_{x \to \infty } {{\int_x^\infty  {{e^{ - {{{t^2}} \over 2}}}} dt} \over {{e^{ - {{{x^2}} \over 2}}}}} = \mathop {\lim }\limits_{x \to \infty } {{ - {e^{ - {{{x^2}} \over 2}}}} \over { - x{e^{ - {{{x^2}} \over 2}}}}} \to 0$$
Applying L'hospital's Rule ,
$$\mathop {\lim }\limits_{x \to \infty } {{x\int_x^\infty  {{e^{ - {{{t^2}} \over 2}}}} dt} \over {{e^{ - {{{x^2}} \over 2}}}}} = \mathop {\lim }\limits_{x \to \infty } {{\int_x^\infty  {{e^{ - {{{t^2}} \over 2}}}} dt - x{e^{ - {{{x^2}} \over 2}}}} \over { - x{e^{ - {{{x^2}} \over 2}}}}} \\= \mathop {\lim }\limits_{x \to \infty } {{ - {e^{ - {{{x^2}} \over 2}}}} \over { - \left( {{e^{ - {{{x^2}} \over 2}}} - {x^2}{e^{ - {{{x^2}} \over 2}}}} \right)}} + 1 = \mathop {\lim }\limits_{x \to \infty } {1 \over {1 - {x^2}}} + 1 = 0 $$
A: $$x e^{x^2/2}\int_{x}^{+\infty}e^{-t^2/2}\,dt = x\int_{x}^{+\infty}e^{(x-t)(x+t)/2}\,dt=x\int_{0}^{+\infty}e^{-t(2x+t)/2}\,dt=\int_{0}^{+\infty}e^{-\frac{t^2}{2x^2}}e^{-t}\,dt$$
By the dominated convergence theorem, the limit of the ratio is so:
$$\int_{0}^{+\infty}e^{-t}\,dt = 1.$$
