limit of and integral depending on n Can somebody give me some tips about how can find the next limit, please?
$$\ \lim_{n\to \infty} \int_{0}^4 \sqrt[n]{x^n+(4-x)^n} dx $$I found that $$\int_{0}^4 \sqrt[n]{x^n+(4-x)^n} =2\int_{0}^2 \sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.
 A: Note that the p-norm
of a 2D vector is defined as
$$
\left\| {\bf x} \right\|_{\,p}  = \left( {x_{\,1} ^{\,p}  + x_{\,2} ^{\,p} } \right)^{\,1/p} 
$$
and it is known that
$$
\mathop {\lim }\limits_{p\, \to \,\infty } \left\| {\bf x} \right\|_{\,p}  = \left\| {\bf x} \right\|_{\,\infty }  = \max \left\{ {\left| {x_{\,1} } \right|,\left| {x_{\,2} } \right|} \right\}
$$
Therefore
$$
\eqalign{
  & \mathop {\lim }\limits_{n\, \to \,\infty } \,\;\int_{x = 0}^{\,4} {\root n \of {x^{\,n}  + \left( {4 - x} \right)^{\,n} } dx}  =   \cr 
  &  = \,\;\int_{x = 0}^{\,4} {\left( {\mathop {\lim }\limits_{n\, \to \,\infty } \root n \of {x^{\,n}  + \left( {4 - x} \right)^{\,n} } } \right)dx}  =   \cr 
  &  = \,\;\int_{x = 0}^{\,4} {\left( {\max \left\{ {\left| x \right|,\left| {4 - x} \right|} \right\}} \right)dx}  =   \cr 
  &  = \,\;2\int_{x = 0}^{\,2} {\left( {4 - x} \right)dx}  = \,\;2\int_{x = 2}^{\,4} {x\,dx}  = 12 \cr} 
$$
A: Hint: assuming Mark’s comment is correct, 
$2^{1/n}\max(x,4-x) \geq (x^n+(4-x)^n)^{1/n} \geq \max(x,4-x)$
A: NOT A SOLUTION:
Not sure if this will be of help. I would start by repositioning the integrand:
\begin{equation}
I = \int_0^2 \sqrt[n]{x^n + \left(4 - x\right)^n}\:dx = \int_0^2 (4 - x) \cdot \sqrt[n]{\left(\frac{x}{4 - x}\right)^n + 1}\:dx 
\end{equation}
Now let $t = \dfrac{x}{4 - x}$ :
\begin{equation}
I = \int_0^1 \frac{4}{t + 1}\sqrt[n]{t^n + 1}\frac{4}{\left(t + 1\right)^2}\:dt = 16 \int_0^1 \frac{\sqrt[n]{t^n + 1}}{\left(t + 1\right)^3}\:dt
\end{equation}
I will leave it to qualified minds to take it further.  If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit)  is equal to the integral of the limit of the integrand i.e. 
\begin{equation}
 \lim_{n\rightarrow \infty} 16 \int_0^1 \frac{\sqrt[n]{t^n + 1}}{\left(t + 1\right)^3}\:dt = 16 \int_0^1 \left[ \lim_{n\rightarrow \infty} \frac{\sqrt[n]{t^n + 1}}{\left(t + 1\right)^3}\right]\:dt
\end{equation} 
Then, 
\begin{align}
 I = 16 \int_0^1 \left[ \lim_{n\rightarrow \infty} \frac{\sqrt[n]{t^n + 1}}{\left(t + 1\right)^3}\right]\:dt = 16 \int_0^1 \left[ \lim_{n\rightarrow \infty} \frac{t}{\left(t + 1\right)^3}\right]\:dt = \frac{1}{8}
\end{align}
But I am unsure if that is correct or not. As before, I will leave it to more qualified minds. 
