Question

Solve
${lim}_{x\to 0}{\left(\frac{1^x+2^x+3^x+...+n^x}{n}\right)}^{1/2}=$

Answer 1

We have,

$\underset{x\to 0}{lim}{\left(\frac{1^x+2^x+3^x+4^x+-----+n^x}{n}\right)}^{1/2}$

we can write this,

$=e^{\underset{x\to 0}{lim}}\log {\left(\frac{1^x+2^x+3^x+4^x+-----+n^x}{n}\right)}^{1/2}$

$\because e^{\log x=1}$

$=e^{\underset{x\to 0}{lim}}\frac{1}{2}\log \left(\frac{1^x+2^x+3^x+4^x+-----+n^x}{n}\right)$

$=e^{\frac{1}{2}\underset{x\to 0}{lim}}[\frac{1^x+2^x+3^x+4^x+-----+n^x}{n}-1]$

$=e^{\frac{1}{2}\underset{x\to 0}{lim}}\left[\frac{(1^x-1)+(2^x-1)+(3^x-1)+----+(n^x-1)}{n}\right]$

$=e^{\frac{1}{2}\underset{x\to 0}{lim}}\left[\frac{x\log 1+x\log 2+x\log 3+----+x\log n}{n}\right]$

$=e^{\frac{1}{2}\underset{x\to 0}{lim}}\frac{x\log (\mathrm{1.2.3.}-----n)}{n}$

taking limt and we get,

$=e^{\frac{1}{2}}\frac{0\times \log (\mathrm{1.2.3.4}---n)}{n}$

$=e^o=1$

Hence this is the answer.