# finding the limit.

• Oct 23rd 2011, 12:08 PM
hmmmm
finding the limit.
How do I prove that the limit of the function

$\displaystyle limit_{n\rightarrow \infty}(x_1^n+x_2^n+x_3^n+.....+x_k^n)^{(1/n)}$

where $\displaystyle 0<x_1,x_2,.....,x_k\leq 1$ is equal to $\displaystyle max\{x_1,x_2,.....,x_k\}$

thanks for any help

solution

using the sandwich theorem:

denoting $\displaystyle x_m=max\{x_1,x_2,.....,x_k\}$ we have

$\displaystyle \{x_1^n+x_2^n+.....x_k^n\}^{1/n} =x_m\{(\frac{x_1}{x_m}^n+.....+1+....(\frac{x_k}{x _m})^n\}\leq x_m(k)^{(1/n)}$

and $\displaystyle Limit_{n\rightarrow\infty}x_m(k)^{(1/n)}=x_m$ (this is the upper bound)

$\displaystyle x_m\{(\frac{x_1}{x_m}^n+.....+1+....(\frac{x_k}{x_ m})^n\}\geq x_m$

and $\displaystyle Limit_{n\rightarrow\infty}x_m=x_m$

so by the sandwich theorem we have that $\displaystyle limit_{n\rightarrow \infty}(x_1^n+x_2^n+x_3^n+.....+x_k^n)^(1/n)=x_m$

Solved this after I posted it (I think)
• Oct 23rd 2011, 01:02 PM
SammyS
Re: finding the limit.
Quote:

Originally Posted by hmmmm
How do I prove that the limit of the function

$\displaystyle \lim_{n\rightarrow \infty}(x_1^n+x_2^n+x_3^n+.....+x_k^n)^{(1/n)}$

where $\displaystyle x_1,x_2,.....,x_k\leq 1$ is equal to $\displaystyle max\{x_1,x_2,.....,x_k\}$

thanks for any help

solution

using the sandwich theorem:

denoting $\displaystyle x_m=max\{x_1,x_2,.....,x_k\}$ we have

$\displaystyle \{x_1^n+x_2^n+.....x_k^n\}^{1/n} =x_m\{(\frac{x_1}{x_m}^n+.....+1+....(\frac{x_k}{x _m})^n\}\leq x_m(k)^{(1/n)}$

and $\displaystyle \lim_{n\rightarrow\infty}x_m(k)^{(1/n)}=x_m$ (this is the upper bound)

$\displaystyle x_m\{(\frac{x_1}{x_m}^n+.....+1+....(\frac{x_k}{x_ m})^n\}\geq x_m$

and $\displaystyle \lim_{n\rightarrow\infty}x_m=x_m$

so by the sandwich theorem we have that $\displaystyle limit_{n\rightarrow \infty}(x_1^n+x_2^n+x_3^n+.....+x_k^n)^(1/n)=x_m$

Solved this after I posted it (I think)

Is $\displaystyle 0 < x_1,x_2,.....,x_k$?
• Oct 23rd 2011, 01:23 PM
hmmmm
Re: finding the limit.
yes sorry I just missed that out, will edit it now thanks very much