Prove that $\lim_{n \to\infty} \int_0^1 \frac{x^n}{\sqrt{1+x^n}}\, \mathrm dx=0$. 
Prove that $$\lim_{n\to\infty} \int_0^1 \frac{x^n}{\sqrt{1+x^n}}\, \mathrm dx=0.$$

For full disclosure, this is for a homework problem for a real analysis class. I am stuck even figuring out how to approach this problem.  The only thing that comes to mind is that the derivative of $\arctan(x)$, however I highly doubt that is useful.  
 A: For each positive integer $n$, let $$f_n(x) = \frac{x^n}{\sqrt{1+x^n}},\ x\in[0,1].$$ Then for all $n$ and $x$, $$|f_n(x)|=\frac{x^n}{\sqrt{1+x^n}}\leqslant x^n\leqslant 1. $$
It follows from dominated convergence that
$$\lim_{n\to\infty}\int_0^1 \frac{x^n}{\sqrt{1+x^n}}\,\mathsf dx = \int_0^1 \lim_{n\to\infty} \frac{x^n}{\sqrt{1+x^n}}\,\mathsf dx.$$
For $x\in(0,1)$, we have $$\lim_{n\to\infty} f_n(x) = 0, $$
and hence
$$\int_0^1 \lim_{n\to\infty} f_n(x)\,\mathsf dx =0.$$
A: Observe that
$$ 0\leq \frac{x^n}{\sqrt{1+x^n}}\leq x^n$$ for all $x\in[0,1]$, hence
$$ 0\leq \int_0^1\frac{x^n}{\sqrt{1+x^n}}\;dx\leq \int_0^1x^n\;dx=\frac{1}{n+1}$$
for each $n\geq 1$.
A: carmichael561's solution is the most efficient way to solve this problem. Here is a much less efficient solution.
Using the integral defintion (analytically continued) of Gauss's hypergeometric function
\begin{equation}
{}_{2}\mathrm{F}_{1}(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int\limits_{0}^{1} t^{a-1} (1-t)^{c-b-1} (1-zt)^{-a} dt
\end{equation}
for $\mathrm{Re}\,c \gt \mathrm{Re}\,b \gt 0,\,\,|\mathrm{arg}(1-z)| \lt \pi$
We have, using the substitution $y=x^{n}$
\begin{align}
\int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx 
&= \frac{1}{n} \int\limits_{0}^{1} \frac{y^{1/n}}{\sqrt{1+y}} dy \\
&= \frac{\Gamma(1+1/n)\Gamma(1)}{n\Gamma(2+1/n)} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right) \\
\tag{1}
&= \frac{1}{n+1} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right)
\end{align}
Taking the limit of the hypergeometric function and applying the integral definition again yields
\begin{align}
\lim_{n \to \infty} {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right)
&= {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1;2;-1 \right) \\
&= \frac{\Gamma(2)}{\Gamma(1)\Gamma(1)} \int\limits_{0}^{1} \frac{1}{\sqrt{1+t}} dt \\
&= 2(\sqrt{2} - 1)
\end{align}
Substituting this result into equation 1, we have
\begin{equation}
\lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0*2(\sqrt{2} - 1) = 0
\end{equation}
