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Prove that:

$$\lim_{n\to\infty} \sqrt[n]{a^{n} + b^{n}} = \max(a, b)$$

I don't have a clue even how to start the proof here. Any hint is appreciated!

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    $\begingroup$ factor out one of the two varible, assume it is the bigger one of the two. $\endgroup$
    – jimjim
    Commented Sep 13, 2016 at 13:28
  • $\begingroup$ How would you go through this proof if we had $a = 2$ and $b = 3$? Apply the same principle for arbitrary $a$ and $b$. $\endgroup$ Commented Sep 13, 2016 at 13:30

3 Answers 3

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Assume $a\geq b\geq0$ and $a>0$ (if $a=b=0$ the statement is trivial). Write $$ \lim_{n\to\infty} \sqrt[n]{a^{n} + b^{n}} = \lim_{n\to\infty}\left( a\cdot\sqrt[n]{1 + (b/a)^{n}} \right), $$ with $1\geq (b/a)\geq (b/a)^n$. Then observe that $$ 1\leq\sqrt[n]{1 + (b/a)^{n}}\leq\sqrt[n]{2}\leq 1+1/n, $$ and conclude that the limit is exactly $a$.

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Without loss of generality, assume $a \geq b \geq 0$, then we have the following inequality: $$\sqrt[n]{a^n} \leq \sqrt[n]{a^n + b^n} \leq \sqrt[n]{a^n + a^n}.$$ Now apply the squeeze principle and the fact $\lim_{n \to \infty} 2^{1/n} = 1$.

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Hint: Whenever $x>y>0$, $x^n>>y^n$, so the larger term dominates inside the radical, and you can "ignore" the smaller term for large enough $n$ and evaluate easily.

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