Proving $\lim_{p\to\infty} $ $\left\lVert x \right\rVert_p = \left\lVert x \right\rVert_\infty$ How can one prove, that 
$\lim_{p\to\infty} $ $\left\lVert x \right\rVert_P = \left\lVert x \right\rVert_\infty$ applies to all $x \in \mathbb{R^n}$ ? 
I know that two norms$\left\lVert \cdot \right\rVert_a$ and $\left\lVert \cdot \right\rVert_b$ in $\mathbb{R^n}$ are equivalent, if there are constants $c_1,c_2 > 0 $ so that for all $x \in \mathbb{R^n}$ there is an inequation chain 
$c_1 \left\lVert x \right\rVert_a \leq \left\lVert x \right\rVert_b \leq c_2 \left\lVert x \right\rVert_a$
I think I have to use the inequation above somehow to prove the former, yet I don't know how
 A: We have
\begin{align}
\|x\|_p
=&
\left[|x_1|^p+\ldots+|x_i|^p+\ldots+|x_n|^p\right]^{\frac{1}{p}}
\\
=&
\left[\displaystyle\max_{1\leq k\leq n}{|x_k|^p}
\left[
\left(
\frac{|x_1|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}}
\right)^p
+\ldots+
\left(
\frac{|x_i|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}}
\right)^p
+\ldots+
\left(
\frac{|x_n|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}}
\right)^p
\right]
\right]^{\frac{1}{p}}
\\
=&
\displaystyle\max_{1\leq k\leq n}{|x_k|}
\left[
\left(
\frac{|x_1|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}}
\right)^p
+\ldots+
\left(
\frac{|x_i|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}}
\right)^p
+\ldots+
\left(
\frac{|x_n|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}}
\right)^p
\right]^{\frac{1}{p}}
\end{align}
Using a shorter notation $\|x\|_\infty=\displaystyle\max_{1\leq k\leq n}{|x_k|}$, 
$$
\|x\|_p= \|x\|_\infty
\left[
\left(
\frac{|x_1|}{\|x\|_\infty}
\right)^p
+\ldots+
\left(
\frac{|x_i|}{\|x\|_\infty}
\right)^p
+\ldots+
\left(
\frac{|x_n|}{\|x\|_\infty}
\right)^p
\right]^{\frac{1}{p}}
$$
We have two implications.
First.
$$
0\leq \left(\frac{|x_1|}{\|x\|_\infty}\right)^p\leq 1,
\ldots,
0\leq \left(\frac{|x_i|}{\|x\|_\infty}\right)^p\leq 1,
\ldots
0\leq \left(\frac{|x_n|}{\|x\|_\infty}\right)^p\leq 1,
$$
implies 
$$
\|x\|_p\leq \|x\|_\infty\sqrt[p]{n} \quad (\ast)
$$
Second.
$$
1\leq \left[
\left(
\frac{|x_1|}{\|x\|_\infty}
\right)^p
+\ldots+
\left(
\frac{|x_i|}{\|x\|_\infty}
\right)^p
+\ldots+
\left(
\frac{|x_n|}{\|x\|_\infty}
\right)^p
\right]^{\frac{1}{p}}
$$
implies 
$$
\|x\|_\infty\leq \|x\|_{p}\quad (\ast\ast)
$$
Putting the Inequalities $(\ast)$ and $(\ast\ast)$ together we have 
$$
\|x\|_\infty\leq \|x\|_{p}\leq \|x\|_\infty\cdot \sqrt[p]{n}
$$
Once $\lim_{p\to\infty}\sqrt[p]{n}=1$ for all $n\in\mathbb{N}-\{0\}$ we have 
$$
\lim_{p\to \infty} \|x\|_{p}=\|x\|_\infty
$$
