Question

limit as x tends to infinity of sqrt(x^(1/x))

Answer 1

\[\lim_{x \rightarrow \infty} \sqrt{x^\frac{ 1 }{ x }}\]Show working.

Answer 2

Suppose
\[L=\lim_{x\to\infty}\sqrt{x^{1/x}}\]
Then
\[\ln L=\ln\lim_{x\to\infty}\sqrt{x^{1/x}}~~\Rightarrow~~\ln L=\lim_{x\to\infty}\ln\sqrt{x^{1/x}}\]
Since \(\sqrt{x^{1/x}}=\left(x^{1/x}\right)^{1/2}={\large x^{1/(2x)}}\), and using some log properties, you have
\[\ln L=\lim_{x\to\infty}\frac{\ln x}{2x}=\frac{\infty}{\infty}\]
Apply L'Hopital's rule as many times as necessary.

Answer 3

Thanks!

Answer 4

yw