Proof $||\vec x||_\infty = \max\{|x_i|\}$ I know that the norm of vector $x$ is defined to be $||x||_n=\sqrt[n]{\sum_{i=1}^m|x_i|^n}$ 
How using that i can show that the inf norm is max?
 A: Hint:$$\lim_{n \to \infty}\sqrt[n]{2^n+3^n+5^n}=\\\lim_{n \to \infty}\sqrt[n]{5^n((\frac 25)^n+(\frac 35)^n+1)} \to 5$$You can facotr $max\{|x_i|\}$ in radical ,to proove what you need 
$$||\vec x||_\infty=\lim_{n \to \infty}\sqrt[n]{\sum_{i=1}^m|x_i|^n} $$take $max\{|x_i|\}=M$ so $M \geq \underbrace{|x_i|}_{i=1,2,3,...m}$
now
$$\lim_{n \to \infty}\sqrt[n]{\sum_{i=1}^m|x_i|^n}=\\\lim_{n \to \infty}\sqrt[n]{|x_1|^n+|x_2|^n+...+M^n+...+|X_m|^n}=\\
\lim_{n \to \infty}M\sqrt[n]{(\frac{|x_1|}{M})^n+(\frac{|x_2|}{M})^n+...+(\frac{M}{M})^n+...+(\frac{|x_m|}{M})^n}=\\
M\sqrt[n]{0+0+...+1+0+0+..}=\\M=max\{|x_i|\}$$
$\bf{Remark}:$If two or more of $|x_i|$ are equal to M we have the same result, let's say that $k\in\mathbb{N}$ is the number of parts that are equal to M$$\lim_{n \to \infty}\sqrt[n]{\sum_{i=1}^m|x_i|^n}=\\\lim_{n \to \infty}\sqrt[n]{|x_1|^n+|x_2|^n+...+M^n+...+M^n+...+|X_m|^n}=\\
\lim_{n \to \infty}M\sqrt[n]{(\frac{|x_1|}{M})^n+(\frac{|x_2|}{M})^n+...+(\frac{M}{M})^n+...+(\frac{M}{M})^n+...+(\frac{|x_m|}{M})^n}=\\
M\lim_{n \to \infty}\sqrt[n]{0+0+...1+...+1+0+0+..}=\\M\lim_{n \to \infty}\sqrt[n]{k}=\\
M\lim_{n \to \infty}k^{\frac 1n}\to M\times 1\\M=max\{|x_i|\}$$
