Find $L=\lim \limits_{n\to \infty}\sqrt[n]{x^n+y^n+z^n}$

Question

Find the limit following:

$$L=\lim \limits_{n\to \infty}\sqrt[n]{x^n+y^n+z^n}$$

With $x,\: y\: z\in R$

P.S

I think this limit result is $L=max\left\{x,\: y\: z \right\}$. But i'm not find it, so expect people to help me find out the results by some solution

Extend this limit:

$$S=\lim \limits_{n\to \infty}\sqrt[n]{\sum_{i}^{m}a_{i}}$$ with $a_i\in R,\: i=1,\: m$

Minimum? Looks like it is the one with the maximum absolute value. — Aryabhata, Nov 27 '13 at 01:19
In addition to Aryabhata's comment, if $x,y,z$ are negative, then the limit does not exist. — , Nov 27 '13 at 01:22
For reference: the thing you are describing is called the infinity norm (that is, the limit of the $p$-norms as $p \to \infty$). As the answers below might suggest, it is also called the "maximum norm". — senshin, Nov 27 '13 at 07:32
https://math.stackexchange.com/questions/80340/convergence-of-sqrtnxnyn-for-x-y-0?s=1|145.5010 — Guy Fsone, Jan 29 '18 at 17:03
Possible duplicate of The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. — Guy Fsone, Jan 29 '18 at 17:05

score 15 · Answer 1 · answered Nov 27 '13 at 01:25

15

Put $M = \max(x,y,z).$ Then you have $M^n \le x^n + y^n + z^n \le 3M^n$. Take $n$th roots to ge $$M \le \root{n}\of{x^n + y^n + z^n} \le 3^{1/n}M.$$ Let $n\to \infty$ and see your limit is $M$.

answered Nov 27 '13 at 01:25

ncmathsadist

50,127

I am assuming these constants are nonnegative to avoid negative root issues. – ncmathsadist Nov 27 '13 at 01:26

score 2 · Answer 2 · answered Nov 27 '13 at 01:23

2

The answer should be the max over all terms.

Think of the term which will start dominating the sum when n becomes arbitrarily large ...

answered Nov 27 '13 at 01:23

shyamupa

421

Find $L=\lim \limits_{n\to \infty}\sqrt[n]{x^n+y^n+z^n}$

2 Answers2

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