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. 
 A: 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 $\varepsilon\to0$.
A: 
Herein, we present a way forward that relies on only integration by parts and the squeeze theorem.  To that end we proceed.


Let $I_n(a)$ be the sequence of functions of $a$ represented by the integral 
$$\begin{align}
\bbox[5px,border:2px solid #C0A000]{I_n(a)=\int_a^\infty e^{-nx^2/2}\,dx}\tag 1
\end{align}$$
Integrating by parts the integral on the right-hand side of $(1)$ with $u=-\frac1{nx}$ and $v=e^{-nx^2/2}$ yields
$$\begin{align}
I_n(a)&=\frac{e^{-na^2/2}}{na}-\int_{a}^\infty \frac{e^{-nx^2/2}}{nx^2}\,dx \\\\
&=\frac{e^{-na^2/2}}{na}\left(1-a\int_a^\infty \frac{e^{-n(x^2-a^2)/2}}{x^2}\,dx\right)\tag 2
\end{align}$$
The term in parentheses on the right-hand side of $(2)$ satisfies the bounds
$$1-\frac{1}{na^2}\le 1-a\int_a^\infty \frac{e^{-n(x^2-a^2)/2}}{x^2}\,dx\le 1 \tag 3$$
Therefore, using $(3)$ reveals
$$\frac{e^{-a^2/2}}{\left(na\right)^{1/n}}\left(1-\frac{1}{na^2}\right)^{1/n}\le I_n(a)^{1/n}\le \frac{e^{-a^2/2}}{\left(na\right)^{1/n}}\tag 4$$
whereupon applying the squeeze theorem to $(4)$ we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\int_a^\infty e^{-nx^2/2}\,dx\right)^{1/n}=e^{-a^2/2}}$$

A: Substituting $x\mapsto\sqrt{\frac{2x}n}$ gives
$$
\int_a^\infty e^{-nx^2/2}\,\mathrm{d}x
=\frac1{\sqrt{2n}}\int_{na^2/2}^\infty e^{-x}\,\frac{\mathrm{d}x}{\sqrt{x}}\tag{1}
$$
For $x\ge na^2/2$, $\frac1{\sqrt{x}}\le\sqrt{\frac2n}\frac1a$. Therefore,
$$
\begin{align}
\frac1{\sqrt{2n}}\int_{na^2/2}^\infty e^{-x}\,\frac{\mathrm{d}x}{\sqrt{x}}
&\le\frac1{na}\int_{na^2/2}^\infty e^{-x}\,\mathrm{d}x\\
&=\frac1{na}e^{-na^2/2}\tag{2}
\end{align}
$$
Since $e^x\ge x$, we get $\sqrt{\frac2n}e^{-x/n}\le\frac1{\sqrt{x}}$. Therefore,
$$
\begin{align}
\frac1{\sqrt{2n}}\int_{na^2/2}^\infty e^{-x}\,\frac{\mathrm{d}x}{\sqrt{x}}
&\ge\frac1n\int_{na^2/2}^\infty e^{-x(1+1/n)}\,\mathrm{d}x\\
&=\frac1{n+1}e^{-(n+1)a^2/2}\tag{3}
\end{align}
$$
Combining $(1)$, $(2)$, and $(3)$ shows that
$$
\left(\frac{e^{-a^2/2}}{n+1}\right)^{1/n}e^{-a^2/2}
\le\left(\int_a^\infty e^{-nx^2/2}\,\mathrm{d}x\right)^{1/n}
\le\left(\frac1{na}\right)^{1/n}e^{-a^2/2}\tag{4}
$$
Applying the Squeeze Theorem to $(4)$ yields
$$
\lim_{n\to\infty}\left(\int_a^\infty e^{-nx^2/2}\,\mathrm{d}x\right)^{1/n}=e^{-a^2/2}\tag{5}
$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty}\pars{\int_{a}^{\infty}\expo{-nx^{2}/2}\dd x}^{1/n}:\
?.\qquad a >0}$.

Lets consider
\begin{align}
&\bbox[10px,border:0.1em groove navy]{\lim_{n \to \infty}
\bracks{{1 \over n}\,\ln\pars{\int_{a}^{\infty}\expo{-nx^{2}/2}\dd x}}} =
\lim_{n \to \infty}
\bracks{{1 \over n}\,\ln\pars{{2 \over \root{n}}\int_{a\root{n/2}}^{\infty}\expo{-x^{2}}\dd x}}
\\[5mm] = &\
\lim_{n \to \infty}
\bracks{{1 \over n}\,\ln\pars{\int_{a\root{n/2}}^{\infty}\expo{-x^{2}}\dd x}}
\\[5mm] = &\
\bbox[10px,border:0.1em groove navy]{\lim_{n \to \infty}
\bracks{\ln\pars{\int_{a\root{\bracks{n + 1}/2}}^{\infty}\expo{-x^{2}}\,\dd x} -
\ln\pars{\int_{a\root{n/2}}^{\infty}\expo{-x^{2}}\dd x}}}\label{1}\tag{1}
\\
& \pars{~\mbox{The last expression is found with}\
Stoltz\!-\!Ces\grave{a}ro\ Theorem~}
\end{align}

Note that $\ds{\pars{~\mbox{with the}\ \Lambda > 0~}}$:

\begin{align}
&\int_{\Lambda}^{\infty}\expo{-x^{2}}\dd x =
{\expo{-\Lambda^{2}} \over 2\Lambda} -
{1 \over 2}\int_{\Lambda}^{\infty}{\expo{-x^{2}} \over x^{2}}\,\dd x
\\[1cm] & \mbox{and}\
\lim_{\Lambda \to \infty}\bracks{\int_{\Lambda}^{\infty}\expo{-x^{2}}\dd x -
{\expo{-\Lambda^{2}} \over 2\Lambda}}  = 0
\ \mbox{because}\quad
0 < \verts{{1 \over 2}\int_{\Lambda}^{\infty}{\expo{-x^{2}} \over x^{2}}\,\dd x}
< {\expo{-\Lambda^{2}} \over 2\Lambda}
\stackrel{\Lambda \to \infty}{\to} 0\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}:
\begin{align}
&\bbox[10px,border:0.1em groove navy]{\lim_{n \to \infty}
\bracks{{1 \over n}\,\ln\pars{\int_{a}^{\infty}\expo{-nx^{2}/2}\dd x}}}
\\[5mm] = &
\lim_{n \to \infty}\braces{\bracks{-\,{a^{2}\pars{n + 1} \over 2} -
\ln\pars{\root{2}a} - {1 \over 2}\,\ln\pars{n + 1}} -
\bracks{-\,{a^{2}n \over 2} - \ln\pars{\root{2}a} -
{1 \over 2}\,\ln\pars{n}}}
\\[5mm] = &\
\bbox[10px,border:0.1em groove navy]{-\,{a^{2} \over 2}}
\end{align}

Then,
$$
\color{#f00}{%
\lim_{n \to \infty}\pars{\int_{a}^{\infty}\expo{-nx^{2}/2}\dd x}^{1/n}} =
\color{#f00}{\expo{-a^{2}/2}}
$$
