mathop {lim }limits_{x to 0^+} {x^x} = | vedclass

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Question

(B) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\log y = x \log x$.Now,we evaluate the limit as $x 	o 0^+$:$\mathop {\lim }\lim

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(B) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\log y = x \log x$.Now,we evaluate the limit as $x 	o 0^+$:$\mathop {\lim }\limits_{x 	o 0^+} \log y = \mathop {\lim }\limits_{x 	o 0^+} x \log x = \mathop {\lim }\limits_{x 	o 0^+} \frac{\log x}{1/x}$.Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator:$= \mathop {\lim }\limits_{x 	o 0^+} \frac{1/x}{-1/x^2} = \mathop {\lim }\limits_{x 	o 0^+} (-x) = 0$.Since $\log y 	o 0$,it follows that $y 	o e^0 = 1$.Therefore,$\mathop {\lim }\limits_{x 	o 0^+} x^x = 1$.

$\mathop {\lim }\limits_{x \to 0^+} {x^x} = $

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