Computation $\lim_n \left(\int_a^{\infty}e^{-\frac{nx^2}{2}}\,dx\right)^{\frac{1}{n}}$ I would like a hint to compute
$$ \lim_{n\rightarrow\infty} \left(\int_a^{\infty}e^{-\frac{nx^2}{2}}\,dx\right)^{\frac{1}{n}},$$
where $a>0$.
I thought of applying $\exp(\log)$ and then l'Hôpital, but nothing. 

If $f$ is bounded and continuous on a bounded open interval $I\subset\mathbb R$ then one has
$$
\lim_{p\to\infty}\left(\int_I|f(x)|^pdx\right)^{1/p}=\|f\|_\infty:=\sup_{x\in I}|f(x)|.
$$
Our interval isn't bounded but since $e^{-x^2/2}$ decays so quickly we might expect the same result here. We first note that for all $x\ge a$ and $\varepsilon>0$, $e^{-x^2/2}=e^{-x^{2-\varepsilon}/2}e^{-x^{\varepsilon}/2}\le e^{-a^{2-\varepsilon}/2}e^{-x^\varepsilon/2}$. Since $nx^\varepsilon\ge x^{\varepsilon}$ we get
$$
\int_a^\infty e^{-nx^2/2}dx\le e^{-na^{2-\varepsilon}}\int_a^\infty e^{-x^{\varepsilon}/2}dx,
$$
so taking $n^{\text{th}}$ roots and letting $n\to\infty$ we find
$$
\limsup_{n\to\infty}\left(\int_a^\infty e^{-nx^2/2}dx\right)^{1/n}\le e^{-a^{2-\varepsilon}/2}.
$$
On the other hand, there exists $\delta>0$ such that $e^{-x^2/2}\ge e^{-a^2/2}-\varepsilon$ if $x\in[a,a+\delta]$, so
$$
\int_a^\infty e^{-nx^2/2}dx\ge\int_a^{a+\delta}e^{-nx^2/2}dx\ge(e^{-a^2/2}-\varepsilon)^n\delta.
$$
Taking $n^{\text{th}}$ roots and letting $n\to\infty$ again, we find
$$\liminf_{n\to\infty}\left(\int_a^\infty e^{-nx^2/2}dx\right)^{1/n}\ge e^{-a^2/2}-\varepsilon
$$
and the result follows by letting $\varepsilon\to0$.
