I would like to find the following limit for any $x\in\mathbb R^m$:
$$\lim_{n\to\infty} \sqrt[n]{\sum_{i=1}^m |x_i|^n}$$
Is this even possible and if so, how do I start? As I do not even know what $x$ is I am not quite sure where to begin. Any hints?
The answer is the maximum value of $|x_i|$.
The limit for any $n$ is minimal when only one $x_i$ is non-zero, and maximal if all $x_i$ are equal.
The sum over $m$ terms is limited. Work from there as $\lim_{n\to\infty} \sqrt[n]{m} = 1$
Let $M=\max\{|x_1|,\ldots,|x_m|\}$. Then $$ M^n\leq\sum_{i=1}^m|x_i|^n\leq m M^n\implies M\leq\sqrt[n]{\sum_{i=1}^m|x_i|^n}\leq\sqrt[n]{m}M. $$ Now use the Squeeze Theorem and the fact that $\sqrt[n]{m}\to 1$ as $n\to\infty$.
Let $i$ be such that $x_p=\min_{1\le i\le m} |x_i|$ and $j$ be such that $x_q=\max_{1\le i\le m} |x_i|$
Then $\{\sum_{i=1}^m |x_i|^n\}^{\frac{1}{n}}\le (m|x_q|^n)^{\frac{1}{n}}=|x_q|.m^{\frac{1}{n}}$
Also $\{\sum_{i=1}^m |x_i|^n\}^{\frac{1}{n}}\ge m|x_p|^n)^{\frac{1}{n}}=|x_p|.m^{\frac{1}{n}}$
Now for any $c>0$ ;$\lim c^{\frac{1}{n}}=1$
Hence $\lim_{n\to \infty} m^{\frac{1}{n}}=1$
So $x_p\le\lim_{n\to \infty}\{\sum_{i=1}^m |x_i|^n\}^{\frac{1}{n}}\le x_q$
