Question

$\underset{n\to \infty }{lim}{n}^2(x^{1/n}-x^{1/n+1}),x>0$, is equal to

• A. $0$
• B. $e^x$
• C. ${\log }_{e}x$
• D. None of these

Answer 1

The correct option is C ${\log }_{e}x$

$\underset{n\to \infty }{lim}{n}^2(x^{\frac{1}{n}}-x^{\frac{1}{n+1}})=\underset{n\to \infty }{lim}{n}^2.x^{\frac{1}{n+1}}(x^{\frac{1}{n}-\frac{1}{n+1}}-1)=\underset{n\to \infty }{lim}{x}^{\frac{1}{n+1}}(x^{\frac{1}{n(n+1)}}-1)n^2=\underset{n\to \infty }{lim}{x}^{\frac{1}{n+1}}.\frac{x^{\frac{1}{n(n+1)}}-1}{\frac{1}{n(n+1)}}.\frac{n^2}{n(n+1)}=1.{\log }_{e}x.1={\log }_{e}x$
Hence, option 'C' is correct.