How to evaluate $\lim\limits_{p \rightarrow \infty} \left(\sum_\limits{i=1}^n \left|x_i-y_i\right|^p\right)^{\frac{1}{p}}$

Question

I'd like to know why $\lim\limits_{p \rightarrow \infty} \left(\sum_\limits{i=1}^n \left|x_i-y_i\right|^p\right)^{\frac{1}{p}} = \max\limits_{1\le i \le n} \left| x_i-y_i\right|$ for $\mathbf{x},\mathbf{y}\in \mathbb{R}^n$.

So I started by checking a simpler expression:

$\lim\limits_{x\rightarrow \infty} ((6-3)^x+(5-1)^x)^{\frac{1}{x}}=4$

I don't know how to get 4. The expression inside the parenthesis is indeterminate $(\infty + \infty)$ and I don't know of any way to rewrite it so that I can remove the exponents.

score 5 · Accepted Answer · answered Feb 06 '15 at 11:55

$$ \lim_{x\to \infty} ((6-3)^x+(5-1)^x)^{1/x}=\lim_{x\to \infty} (3^x+4^x)^{1/x}=\lim_{x\to \infty}4\Bigl(\Bigl(\frac{3}{4}\Bigr)^x+1\Bigr)^{1/x}=4. $$ For the general case, let $a_i=|x_i-y_i|$ and assume without loss of generality that $\max a_i=a_1>0$. Then $$ \Bigl(\sum_{i=1}^na_i^p\Bigr)^{1/p}=a_1\Bigl(\sum_{i=1}^n\Bigl(\frac{a_i}{a_1}\Bigr)^p\Bigr)^{1/p}. $$ Since $$ 1\le\sum_{i=1}^n\Bigl(\frac{a_i}{a_1}\Bigr)^p\le n, $$ $$ \lim_{p\to\infty}\Bigl(\sum_{i=1}^n\Bigl(\frac{a_i}{a_1}\Bigr)^p\Bigr)^{1/p}=1. $$

How to evaluate $\lim\limits_{p \rightarrow \infty} \left(\sum_\limits{i=1}^n \left|x_i-y_i\right|^p\right)^{\frac{1}{p}}$

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