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The value of \lim _{x \rightarrow 0} x^{x} is

Option: 1

0


Option: 2

1


Option: 3

e


Option: 4

None of these 


Answers (1)

best_answer

The given limit is of 0^0 form

\\\text { Let } y=x^{x} \\ \text{Apply log both side}\\ \Rightarrow \log y=x \log x \\ \Rightarrow \lim _{y \rightarrow 0} \log y=\lim _{x \rightarrow 0} x \log x\\ \Rightarrow \lim _{y \rightarrow 0} \log y=\lim _{x \rightarrow 0}\frac{log(x)}{\frac{1}{x}}\\L'H\,\,Rule \\\Rightarrow \lim _{y \rightarrow 0} \log y=\lim _{x \rightarrow 0}\frac{\frac{1}{x}}{-\frac{1}{x^2}} \\\Rightarrow \lim _{y \rightarrow 0} \log y=\lim _{x \rightarrow 0}-x \\\Rightarrow \lim _{y \rightarrow 0} \log y=0\\\Rightarrow \lim _{x \rightarrow 0} x^{x}=1

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