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. 


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

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

