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\mathrm{\lim _{x \rightarrow \infty} x^{1 / x} \text { is }}

Option: 1

\infty


Option: 2

0


Option: 3

1


Option: 4

not defined


Answers (1)

best_answer

Let                     \mathrm{K=\lim _{x \rightarrow \infty} x^{\frac{1}{x}}}

Taking logarithms on both sides

                \mathrm{\log K=\lim _{x \rightarrow \infty} \log \left[x^{1 / x}\right]}

                \mathrm{\log K=\lim _{x \rightarrow \infty}\left\{\frac{1}{x} \log x\right\}\left(\frac{\infty}{\infty} \text { form }\right)}

                \mathrm{\log K=\lim _{x \rightarrow \infty}\left(\frac{1 / x}{1}\right)}

                                                        \mathrm{\text { (applying L.H. Rule) }}

                   \mathrm{\log K=0}

\therefore                      \mathrm{K=e^0=1}

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Nehul

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