If $\displaystyle f(x) = \lim_{n \to \infty} ((\cos x )^n + (\sin x)^n))^{1/n}$. Then $_0 ^{\pi /2}\int f(x).dx$ is? 
$\displaystyle f(x) = \lim_{n \to \infty} ((\cos x )^n + (\sin x)^n))^{1/n}$. Then $\int _0 ^{\pi /2} f(x)\,dx$ is?    

I proceeded by attempting to solve the limit first as the integral for the function with the powers is very difficult to solve but even using l'hospital twice (after taking log) I am not able to resolve the limit. How do I proceed?
 A: $\lim (a^{n}+b^{n})^{1/n}=\max \{a,b\}$ for any $a , b>0$ which makes the question very easy to answer. To prove this take $a<b$ and write $(a^{n}+b^{n})^{1/n}=a(1+(b/a)^{n})^{1/n}$; use the fact that $\log(1+x)$ behaves like $x$ for $x$ small. 
A: For numbers $a,b \ge 0$ we have
$(a^n+b^n)^{1/n} \to \max \{a,b\}$ as $n \to \infty.$. Hence
$f(x)= \max \{ \cos x, \sin x\}$ for $x \in [0, \pi/2].$ This gives
$f(x)= \cos x$ $x \in [0, \pi/4]$ and $f(x)= \sin x$ for $ x \in [\pi/4,\pi/2].$
A: $$f(x)=\lim_{n\rightarrow \infty}\bigg((\cos x)^n+(\sin x)^n\bigg)^{\frac{1}{n}}$$
$$=\left\{\begin{matrix}
\lim_{n\rightarrow \infty}\cos x\bigg[(\tan x)^n+1\bigg]^{\frac{1}{n}}=\cos x\;, \displaystyle x\in \bigg(0,\frac{\pi}{4}\bigg) & \\\
\lim_{n\rightarrow \infty}\sin x\bigg[(\cot x)^n+1\bigg]^{\frac{1}{n}}=\sin x\;, \displaystyle x\in \bigg(\frac{\pi}{4},\frac{\pi}{2}\bigg)  & 
\end{matrix}\right.$$
$$\int^{\frac{\pi}{2}}_{0}f(x)dx = \int^{\frac{\pi}{4}}_{0}\cos xdx +\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\sin xdx$$
