How does one evaluate $\lim _{n\to \infty }\left(\sqrt[n]{\int _0^1\:\left(1+x^n\right)^ndx}\right)$? I tried using Lebesgue's dominated convergence theorem and I'm getting $\lim _{n\to \infty }\left(1+\int _0^1x^ndx\:\right)$ which is $1$. But the answer should be 2.
 A: Showing that $2$ is an upper bound for the limit is straightforward, since for $0 \leqslant x \leqslant 1$, we have $(1 +x^n)^n \leqslant 2^n$, and
$$ \left(\int_0^1 (1 + x^n)^n \, dx  \right)^{1/n} \leqslant 2, \\ \implies \limsup_{n \to \infty}\left(\int_0^1 (1 + x^n)^n \, dx  \right)^{1/n} \leqslant 2.$$
For a useful lower bound, note that the integrand becomes more and more concentrated in a left neighborhood of $x = 1$ as $n$ increases. If $1 - \delta/n < x \leqslant 1$, then 
$$1 + x^n > 1 +(1 - \delta/n)^n > 1 + 1 - n(\delta/n) = 2 - \delta,$$
and
$$\int_0^1 (1 + x^n)^n \, dx > \int_{1 - \delta/n}^1 (1 + x^n)^n \, dx \\ > \frac{\delta}{n}(2 - \delta)^n.$$
Hence,
$$\liminf_{n \to \infty} \left(\int_0^1 (1 + x^n)^n \, dx  \right)^{1/n} > \liminf_{n \to \infty}\,(2-\delta)\frac{\delta^{1/n}}{n^{1/n}} = 2 - \delta.$$
Since $\delta > 0$ can be arbitrarily small, we have 
$$2 \leqslant \liminf_{n \to \infty} \left(\int_0^1 (1 + x^n)^n \, dx\right)^{1/n}  \leqslant \limsup_{n \to \infty} \left(\int_0^1 (1 + x^n)^n \, dx\right)^{1/n}   \leqslant 2.$$
This shows both that the limit exists and the value is $2$.
A: Let
$$ 
A_n = \Big( \int_{0}^1 (x^n + 1)^n dx \Big) ^{1/n} 
$$
$$ 
A_n^{n} = \int_{0}^1 (x^n + 1)^n dx
$$
$$ 
A_n^{n} = \int_{0}^1 (x^n + 1)^n dx   = \int_{0}^1 \Sigma_{i=0}^{n} \binom{n}{i} x^{ni} dx = \Sigma_{i=0}^{n} \binom{n}{i} \int_{0}^1 x^{ni} dx 
$$
$$
A_n^{n} = \Sigma_{i=0}^{n} \binom{n}{i} \frac{1}{n \cdot i + 1} $$
Hence for the upper bound we have:
$$A_n^{n} \le \Sigma_{i=0}^{n} \binom{n}{i} =2^n $$
$$A_n \le 2 $$
And for the lower bound we have:
$$A_n^{n} \ge \Sigma_{i=0}^{n} \binom{n}{i} \frac{1}{n \cdot n + 1} = \frac{2^n}{n^2+1} \ge \frac{2^n}{(n+1)^2}  $$
$$A_n \ge \frac{2}{(n+1)^{2/n}}  $$
Since
$$\lim_{n\to\infty} \frac{2}{(n+1)^{2/n}} = 2 \cdot \lim_{n\to\infty} (n+1)^{-2/n} = 2 \cdot \lim_{n\to\infty} e^{-ln(n+1)/n } = 2 \cdot  e^{\lim_{n\to\infty} -ln(n+1)/n } =2 \cdot  e^{0}  = 2$$
Therefore
$$\lim_{n\to\infty} A_n \ge 2$$
Hence
$$\lim_{n\to\infty} A_n = 2$$
A: From $x^n \le 1$ for all $x \in [0,1]$, we get
$$\int_0^1(2x^n)^n\,dx\le\int_0^1(1+x^n)^n\,dx \le\int_0^12^n\,dx$$and therefore$$\frac{2^n}{n^2+1}\le\int_0^1(1+x^n)^n\,dx \le2^n$$
Since it is obvious that $\lim _{n\to \infty }\left(\sqrt[n]{\frac{2^n}{n^2+1}}\right)=\lim _{n\to \infty }\left(\sqrt[n]{2^n}\right)=2$, we can deduce that the limit exists and its value is $2$.
A: Let $I_n$ be given by 
$$I_n=\int_0^1(1+x^n)^n\,dx \tag1$$
From the binomial theorem,  we can write 
$$(1+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^{nk}\tag 2$$
Using $(2)$ in $(1)$ reveals
$$\begin{align}
I_n&=\int_0^1 \sum_{k=0}^n\binom{n}{k}x^{nk}\,dx\\\\
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{1+nk}\tag3
\end{align}$$
Clearly from $(3)$, we obtain the estimates for $(I_n)^{1/n}$
$$\frac{2}{(1+n^2)^{1/n}}\le (I_n)^{1/n}\le 2 \tag 4$$
whereupon applying the squeeze theorem to $(4)$ yields the coveted limit
$$\lim_{n\to \infty}\left(\int_0^1(1+x^n)^n\,dx\right)^{1/n}=2$$
as was to be shown!

NOTE:
The bounds given in $(4)$ follow by letting $k=0$ and $k=n$ in the term $\frac{1}{1+nk}$ in the binomial expansion of $(1+x^n)^n$.  Since $x\le 1$ for $x\in [0,1]$, these bounds are tantamount to the bounds $2x^n\le 1+x^n\le 2$ in $(1)$ (as used by @didgogns), which lead immediately to $(4)$.  
A: Let $A_n$ denote our expression. The easy estimate is $A_n\le 2.$ From below we have
$$ A_n > \left (\int_{1-1/n^2}^1(1+x^n)^n\, dx\right )^{1/n} > \left [\left(1+(1-1/n^2)^n\right)^n\cdot (1/n^2)\right ]^{1/n}$$ $$\tag 1 = (1+(1-1/n^2)^n)\cdot 1/n^{2/n}.$$
Now you have two easy limits: $1+(1-1/n^2)^n \to 2$ and $1/n^{2/n} \to 1.$ Thus the limit of $(1)$ is $2$ and we have shown $A_n \to 2$ by the squeeze theorem.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With Laplace Method:

\begin{align}
\int_{0}^{1}\pars{1 + x^{n}}^{n}\,\dd x & =
\int_{0}^{1}\exp\pars{n\ln\pars{1 + \bracks{1 - x}^{n}}}\,\dd x
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\int_{0}^{\infty}\exp\pars{n\ln\pars{2}- {n^{2} \over 2}\, x}\,\dd x =
{2^{n + 1} \over n^{2}}
\end{align}

\begin{align}
&\lim_{n \to \infty}\root[n]{\int_{0}^{1}\pars{1 + x^{n}}^{n}}\,\dd x  =
\lim_{n \to \infty}{{2^{1 + 1/n} \over n^{2/n}}} =
2\exp\pars{\lim_{n \to \infty}\bracks{-\,{2 \over n}\,\ln\pars{n}}}
\\[5mm] =&\
2\exp\pars{\lim_{n \to \infty}
\braces{-2\,{\ln\pars{n + 1} - \ln\pars{n} \over \bracks{n + 1} - n}}}
\label{1}\tag{1}
\\[5mm] = &\
2\exp\pars{\lim_{n \to \infty}
\bracks{-2\,\ln\pars{1 + {1 \over n}}}} = \bbx{2}
\end{align}


Note the Stolz-Ces$\mrm{\grave{a}}$ro Theorem in expression \eqref{1}.

