Enlarge It Then Shrink It

We can use the fact that $(a^n+b^n)^{1/n} \overset{n\to \infty}{\longrightarrow } \max(a,b)$ . Then the the functional sequence $\left\{\left(x^n+(1-x)^n\right)^{1/n}\right\}_n$ converges simply to $\max\{x,1-x\}$ then by the dominated convergence theorem (the integrand is clearly dominated) the limit is $\int_0^1 \max\{x,1-x\} \mathrm{d}x = \frac{3}{4}$

For finite (and convergent) continuous functions on a finite interval, we can always swap the limit and integral (special case of Lebesque's theorem).

Let's evaluate $\lim_n{\left(x^n+(1-x)^n\right)^{1/n}}$ . Due to symmetry, we only need to compute the limit and its integral for $x > 1/2$ and multiply the result by 2.

$\left(x^n+(1-x)^n\right)^{1/n}=x\left(1+\left(\frac{1}{x}-1\right)^n\right)^{1/n}=x\exp{\left(\frac{1}{n}\ln\left(1+\left(\frac{1}{x}-1\right)^n\right)\right)}\,;$ for $1 > x > 1/2$ , $0 < \frac{1}{x}-1 < 1$ , so for sufficiently large $n$ , the term $\left(\frac{1}{x}-1\right)^n$ in the logarithm will be approaching 0 and we can replace the logarithm by the first (linear) term in its Taylor series: $\left(x^n+(1-x)^n\right)^{1/n}\approx x\exp{\left(\frac{1}{n}\left(\frac{1}{x}-1\right)^n\right)}\,.$

Clearly, $\frac{a^n}{n} \rightarrow 0$ for $|a| < 1$ , so we have found the limit to be $x$ in case $x > 1/2$ (and therefore $1-x$ for $x < 1/2$ ).

All that's left is computing $2\int_{1/2}^1{x\mathrm{d}x}=\frac{3}{4}$ .

Due to symmetry, we can compute it for $0<x<0.5$ and multiply the result by two. Then, because $x^n<<1$ and $\lim_{m\rightarrow\infty} (1+ax)^m=1+amx$ , $2\int_0^{0.5} (x^n+(1-x)^n)^{1/n}dx= 2\int_0^{0.5} (1-nx)^{1/n}dx= 2\int_0^{0.5} (1-x)dx=0.75$ .