Question

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

• A. 0
• B. $e^x$
• C. $\ln x$
• D. None of these

Answer 1

Answer: (C)

$\underset{n\to \infty }{\lim }{n}^2\left(x^{1/n}-x^{\frac{1}{n+1}}\right)$
$=\underset{n\to \infty }{\lim }{n}^2\cdot x^{\frac{1}{n+1}}\left(x^{\frac{1}{n}-\frac{1}{n+1}}-1\right)$
$=\underset{n\to \infty }{\lim }{x}^{\frac{1}{n+1}}\left(\frac{1}{x^{n(n+1)}}-1\right)n^2$
$=\underset{n\to \infty }{\lim }{x}^{\frac{1}{n+1}}\cdot \frac{x^{\frac{1}{n(n+1)}}-1}{\frac{1}{n(n+1)}}\cdot \frac{n^2}{n(n+1)}$
$=1\cdot \ln x\cdot 1=\ln x$