<s>[INST] Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$. 
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$. 

My approach: Given that $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx, \forall n\in\mathbb{N}.$ Let us make the substitution $x^n=t$, then $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}.$$
Now since $0\le t\le 1\implies \frac{1}{t}\ge 1\implies \left(\frac{1}{t}\right)^{2/n}\ge 1 \implies 1+\left(\frac{1}{t}\right)^{2/n}\ge 2\implies \sqrt{1+\left(\frac{1}{t}\right)^{2/n}}\ge \sqrt 2.$ 
This implies that $$\frac{1}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le\frac{1}{\sqrt 2}\\ \implies \int_0^1 \frac{dt}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le \int_0^1\frac{dt}{\sqrt 2}=\frac{1}{\sqrt 2}.$$
So, as you can see, I am trying to solve the question using Sandwich theorem. 
Can someone help me to proceed after this? 
Also, in $$\lim_{n\to\infty}nI_n=\lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}},$$ the limit and integral interchangeable?
 [/INST] Yet another approach is to use integration by parts to obtain
\begin{equation*}
I_{n} + I_{n-2} = \int_{0}^{1}x^{n-2}\sqrt{1+x^{2}}\,\mathrm{d}x = \frac{\sqrt{2}}{n-1} - \frac{1}{n-1}I_{n}
\end{equation*}
for $n \geq 2$, from which we get
\begin{equation*}
nI_{n} = \sqrt{2} - (n-1)I_{n-2} = \sqrt{2} - (n-2)I_{n-2} - I_{n-2}. \tag*{(1)}
\end{equation*}
Define $S_{n} = nI_{n}$ for each $n\in \mathbb{N}$. You have already shown that the sequence $\{S_{n}\}_{n\geq 2}$ is bounded above by $1/\sqrt{2}$, and note also that
\begin{align}
S_{n+1} - S_{n} &= \int_{0}^{1}\frac{(n+1)x^{n+1}-nx^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x \\
&\geq \frac{1}{\sqrt{2}}\int_{0}^{1}\big((n+1)x^{n+1} - nx^{n}\big)\,\mathrm{d}x \\
&= \frac{1}{\sqrt{2}(n+1)(n+2)} > 0
\end{align}
so that the sequence $\{S_{n}\}_{n\geq 2}$ is increasing. Thus by the monotone convergence theorem, $S_{n}$ converges to a limit $L$. It is therefore valid to take the limit as $n \to \infty$ on both sides of $(1)$, noting that $I_{n-2} \to 0$, to find that
\begin{equation*}
L=\sqrt{2} - L
\end{equation*}
or
\begin{equation*}
L = \frac{1}{\sqrt{2}}.
\end{equation*}
</s>