Limit question related to integration Find the limit,
$$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$
My try:
$$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx< \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx$$
Now taking the limit I get that,
$$\frac{1}{4}<L<\frac{3}{4}$$
But, how can I get the exact answer!!
This is Problem 11941 from the American Mathematical Monthly.

Substitute 1/2+y for x, y going from -1/2 to 1/2. This a symmetric integral so the positve and negative intervals contribute equally and it suffices to consider only the positve interval. As n goes to infinity (1/2-y)^n is neglible compared to (1/2+y)^n. Hence the root approaches 1/2+ y and the integral approaches 2*Int((1/2+y)*dy) = 1/2+1/4 = 3/4
