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L-lim-x-0-x-x-




Question Number 5821 by FilupSmith last updated on 30/May/16
L=lim_(x→0)  x^x
$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}^{{x}} \\ $$
Answered by bahmanfeshki last updated on 27/Feb/17
lim_(x→0^+ )  e^(xln x) =lim_(x→0^+ )  e^((ln x)/(1/x)) =^(hop) lim_(x→0^+ )  e^((1/x)/(−(1/x^2 ))) =lim_(x→0^+ )  e^(−x) =1
$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{e}^{{x}\mathrm{ln}\:{x}} =\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{e}^{\frac{\mathrm{ln}\:{x}}{\frac{\mathrm{1}}{{x}}}} \overset{{hop}} {=}\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{e}^{\frac{\frac{\mathrm{1}}{{x}}}{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}} =\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{e}^{−{x}} =\mathrm{1} \\ $$

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