TITLE: How does one evaluate $\lim _{n\to \infty }\left(\sqrt[n]{\int _0^1\:\left(1+x^n\right)^ndx}\right)$?
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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 $2$.