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Find \mathrm{\lim _{x \rightarrow \infty} x^{\frac{1}{x}}}

Option: 1

0

 


Option: 2

1


Option: 3

5


Option: 4

3


Answers (1)

firstly, write it as \mathrm{x^{\frac{1}{x}}=e^{\frac{1}{x} \ln x}} now we have such limit

\mathrm{\lim _{x \rightarrow \infty} e^{\frac{\ln x}{x}}=\lim _{x \rightarrow \infty} e^{\frac{f(x)-\infty}{(x)-\infty}}, \text { where } f(x)=\ln x, g(x)=x}

so we can apply L'hospital's rule

\mathrm{\lim _{x \rightarrow \infty} x^{\frac{1}{x}}=\lim _{x \rightarrow \infty} e^{\frac{1}{x} \ln x}=\lim _{x \rightarrow \infty} e^{\frac{1}{x}}=1}

Posted by

Kshitij

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