What is$\lim\limits_{n \rightarrow +\infty} \left(\int_{a}^{b}e^{-nt^2}\text{d}t\right)^{1/n}$? For $\left(a,b\right)) \in \left(\mathbb{R}^{*+}\right)^2$. Let $\left(I_n\right)_{n \in \mathbb{N}}$ be the sequence of improper integrals defined by
$$
\left(\int_{a}^{b}e^{-nt^2}\text{d}t\right)^{1/n}
$$
I'm asked to calculate the limit of $I_n$ when $ \ n \rightarrow +\infty$.
I've shown that
$$
\int_{x}^{+\infty}e^{-t^2}\text{d}t \underset{(+\infty)}{\sim}\frac{e^{-x^2}}{2x}
$$
However, how can I use it ? I wrote that
$$
\int_{a}^{b}e^{-nt^2}\text{d}t=\frac{1}{\sqrt{n}}\int_{\sqrt{n}a}^{\sqrt{n}b}e^{-t^2}\text{d}t
$$
Hence I wanted to split it in two integrals to use two times the equivalent but i cannot sum them so ... Any idea ?
 A: With $f(x)=e^{-x^{2}}$, by letting $M:=f(x_{0})=\max_{x\in[a,b]}f(x)>0$, we have 
\begin{align*}
\int_{a}^{b}f(x)^{n}dx\leq(b-a)M^{n},
\end{align*}
so
\begin{align*}
\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq(b-a)^{1/n}M,
\end{align*}
so
\begin{align*}
\limsup_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq M.
\end{align*}
On the other hand, let $\epsilon>0$ be such that $\epsilon<M$, then there is some open interval $I$ of $x_{0}$ such that $f(x)>M-\epsilon$ for all $x\in I\cap[a,b]$, then 
\begin{align*}
\int_{a}^{b}f(x)^{n}dx\geq\int_{I\cap[a,b]}f(x)^{n}\geq\int_{I\cap[a,b]}(M-\epsilon)^{n}dx=|I\cap[a,b]|(M-\epsilon)^{n},
\end{align*}
so
\begin{align*}
\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\geq|I\cap[a,b]|^{1/n}(M-\epsilon),
\end{align*}
so
\begin{align*}
\liminf_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\geq M-\epsilon,
\end{align*}
now taking $\epsilon\downarrow 0$, then 
\begin{align*}
\liminf_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\geq M.
\end{align*}
In short,
\begin{align*}
M\leq\liminf_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq\limsup_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq M,
\end{align*}
so
\begin{align*}
\lim_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}=M.
\end{align*}
A: Indeed, this case it is easy since by continuity, for any $\varepsilon $ such that $0<\varepsilon< e^{-a^2}$ there exists $d>0$ such that for $t\in (a, a+d)\subset(a,b)$ we have 
$$|e^{-t^2}-e^{-a^2}|\le\varepsilon \implies e^{-t^2}  \ge e^{-a^2}-\varepsilon>0$$ 
 therefore  for all $n$, we get 
$$ (e^{-a^2}-\varepsilon)d^{1/n}\le \left(\int_a^{a+d}e^{-nt^2}dt\right)^{1/n}\le \left(\int_a^be^{-nt^2}dt\right)^{1/n}\le e^{-a^2}$$ you obtain the result by squeezing and letting $\varepsilon \to0$ afthermath. That is $$ e^{-a^2} =\lim_{n\to\infty}\left(\int_a^b e^{-nt^2}dt\right)^{1/n}$$

Generally, It is well known that for continuous function
  $$ \max_{[a,b]}|f| =\lim_{n\to\infty}\left(\int_a^b|f(t)|^ndt\right)^{1/n}$$

A: First answer. This has some problems but now it is fixed.
So you have the result:
\begin{align}\tag{1}
\int^{\infty}_x e^{-t^2}\,dt = \frac{e^{-x^2}}{2x}+o\left(\frac{e^{-x^2}}{x}\right) \ \ \ \text{as} \ \ x\to\infty
\end{align} 
In your last step, you had a mistake. It would be:
\begin{align}
\int^b_a e^{-nt^2}\,dt &= \frac{1}{\sqrt[]{n}}\int^{\sqrt[]{n}b}_{\sqrt[]{n}a}e^{-t^2}\,dt\\
& = \frac{1}{\sqrt[]{n}}\left(\int^{\infty}_{\sqrt[]{n}a}e^{-t^2}\,dt - \int^\infty_{\sqrt[]{n}b}e^{-t^2}\,dt \right)\\
\end{align}
Assume $0<a<b$ $(\star)$. First note that $$\frac{e^{-nb^2}}{n}=o\left(\frac{e^{-na^2}}{n}\right)$$ as $n\to\infty$ (we will avoid writing this from now on). So use $(1)$ to get:
\begin{align}\tag{2}
\int^b_a e^{-nt^2}\,dt  = \frac{e^{-na^2}}{2na}+o\left(\frac{e^{-na^2}}{n}\right)
\end{align}
For $n$ large enough we can take $n$-th root on both sides of $(2)$ to get:
\begin{align}
\left(\int^b_a e^{-nt^2}\,dt\right)^{1/n}&=\left[\frac{e^{-na^2}}{2na}+o\left(\frac{e^{-na^2}}{n}\right)\right]^{1/n}\\
&=e^{-a^2}\frac{1}{n^{1/n}(2a)^{1/n}}\left[1+o\left(1\right)\right]^{1/n}\\
&\to e^{-a^2}
\end{align}
Where we have used $c_n^{1/n}\to 1$ for $c_n$ strictly positive and bounded away from $0$ and the fact that $\sqrt[n]{n}\to 1$.
$(\star)$: If you allow $a=0$, then something similar can be done which is even easier.

Edit One can also come up with the asymptotics of the integral:
\begin{align}
I_n^n=\int^b_a e^{-nt^2}\,dt
\end{align}
Assume $0<a<b$. Note that $t^2$ is monotonically increasing and using The Laplace Method, we get:
\begin{align}
I_n^n\sim \frac{e^{-na^2}}{2an}
\end{align}
Taking $n$-th root we obtain the result:
\begin{align}
\lim_{n\to\infty} I_n = e^{-a^2}
\end{align}
