$\lim \sqrt[n]{a^n + b^n}$ I've seen some answers here for why this limit is the maximum between $a$ and $b$, but all of then included the hypothesis that $a$ and $b$ are both non negative.
It was asked to show that this limit is $\max\{a,b\}$ without any extra hypothesis. Does anyone can give me a hint of how can I do this without using that the numbers are positive? Or the hypothesis is a necessary condition?
I hope you understand the "duplicated' topic
 A: If $a=b=-1$ then the subsequence
$$c_{2n}=\sqrt[2n]{a^{2n}+b^{2n}}=\sqrt[2n]{2}\stackrel{n\to\infty}{\longrightarrow} 1$$
but the max is $-1$, clearly. It's also easy to see that the limit on the odd terms gives
$$\sqrt[2n+1]{a^{2n+1}+b^{2n+1}}=\sqrt[2n+1]{-2}=(-1)\cdot\sqrt[2n+1]{2}\stackrel{n\to\infty}{\longrightarrow} -1$$
so that the limit doesn't exist because your terms are bouncing around.

Addendum: In light of your comment, we can also see why this demonstrates that you really do want both of $a,b\ge 0$. Even only one is insufficient as $a=-b=1$ gives the subsequence
$$c_{2n+1}=\sqrt[2n+1]{a^n+b^n}=\sqrt[2n+1]{0}\stackrel{n\to\infty}{\longrightarrow} 0$$
but the max is clearly $1$, and again, the overall limit doesn't even exist.
A: If $a>b\geq 0$; then $\sqrt[n]{a^n+b^n}=a\sqrt[n]{1+(\frac{b}{a})^n}=ae^{\frac{1}{n}\ln(1+(\frac{b}{a})^n)}$. since $\ln(1+(\frac{b}{a})^n)\sim (\frac{b}{a})^n$ we have $\dfrac{1}{n}\ln(1+(\frac{b}{a})^n)\sim \frac{1}{n}(\frac{b}{a})^n$, hence $\dfrac{1}{n}\ln(1+(\frac{b}{a})^n)\to 0$, and $\sqrt[n]{a^n+b^n}\to a=\max(a,b)$.If  $a=b$ it is easy to show it.
