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  • <img alt="\mathrm{\lim _{x \rightarrow \infty}\left(\frac{x\left(x^{\frac{1}{x}}-1\right)}{\ln x}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Cright

\mathrm{\lim _{x \rightarrow \infty}\left(\frac{x\left(x^{\frac{1}{x}}-1\right)}{\ln x}\right)}
Evaluate

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

5


Answers (1)

\mathrm{}\mathrm{\lim _{x \rightarrow \infty} \frac{x\left(x^{1 / x}-1\right)}{\ln x}=\lim _{x \rightarrow \infty} \frac{e^{\frac{\ln x}{x}}-1}{\frac{\ln x}{x}}=\lim _{y \rightarrow 0^{+}} \frac{e^y-1}{y}=1}
because \mathrm{\frac{\ln x}{x} \rightarrow 0^{+} \text {as } x \rightarrow \infty}

Posted by

Ramraj Saini

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