TITLE: $\displaystyle\lim_{n\to \infty} \sqrt[n]{x_n+y_n}$ is the maximum of the sequences' nth root limits
QUESTION [6 upvotes]: Let $x_n $ and $y_n$ be two sequences of positive real numbers then  $\displaystyle\lim_{n\to \infty} \sqrt[n]{x_n+y_n}=  \max\left\{\lim_{n\to \infty} \sqrt[n]{x_n} \hspace{1 mm} ; \lim_{n\to \infty} \sqrt[n]{y_n} \right\}  $
I showed the result if both $x_n $ and $y_n$ converge to different limits but I don't know what to do in case they are equal(the limits). EDIT:
let $\displaystyle\lim_{n\to \infty} \sqrt[n]{x_n}=l$ and $\displaystyle\lim_{n\to \infty} \sqrt[n]{y_n}=l'$ and $l>l'$ 
we have $|\sqrt[n]{x_n}-l|<\epsilon$ and  $|\sqrt[n]{y_n}-l'|<\epsilon$ take $\epsilon=\displaystyle\frac{l-l'}{2}$ then $l-\displaystyle\frac{l-l'}{2}<\sqrt[n]{x_n}$
$\implies \left(\displaystyle\frac{l+l'}{2}\right)^n<x_n$ we also have $\sqrt[n]{y_n}<l'+\displaystyle\frac{l-l'}{2}$ then ${y_n}<\left(\displaystyle\frac{l+l'}{2}\right)^n$ hence $x_n>y_n$ then $\sqrt[n]x_n <\sqrt[n]{x_n+y_n}< \sqrt[n]{2x_n}$,hence by the squeeze theorem $\sqrt[n]{x_n+y_n}=l$
But what if $l=l'$

REPLY [4 votes]: Suppose 
$$\lim_{n\to\infty}\sqrt[n]{x_n}=\lim_{n\to\infty}\sqrt[n]{y_n}=L\,$$  then:
$$\sqrt[n]{x_n+y_n}=\sqrt[n]{x_n}\;\sqrt[n]{1+\left(\frac{y_n}{x_n}\right)}\xrightarrow[n\to\infty]{} L\cdot 1=L$$
since by the squeeze theorem
$$1\leftarrow1\leq\sqrt[n]{1+\left(\frac{y_n}{x_n}\right)}\leq\sqrt[n] 3\rightarrow 1$$