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.

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.
