Question

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

• A. ${\left(n!\right)}^{\frac{a}{n}}$
• B. ${\left(n!\right)}^{\frac{n}{a}}$
• C. ${\left(n!\right)}^{an}$
• D. None of these

Answer 1

Answer: (A)

$\underset{x\to 0}{\lim }$ ${\left(\frac{1^x+2^x+3^x+.....+n^x}{n}\right)}^{\frac{a}{x}}$ $(1^{\infty }$ form$)$
$=e^{\underset{x\to 0}{\lim }\left(\frac{1^x+2^x+...+n^x}{n}-1\right)\left(\frac{a}{x}\right)}$
$=e^{\underset{x\to 0}{\lim }\left[\left(\frac{1^x-1}{x}\right)+\left(\frac{2^x-1}{x}\right)+...+\left(\frac{n^x-1}{x}\right)\right]\frac{a}{n}}$
$=e^{log\left(1\cdot \mathrm{2...}n\right)\cdot \frac{a}{n}}=e^{log\left(n!\right)\frac{a}{n}}$
$={\left(n!\right)}^{\frac{a}{n}}$