Question Number 150599 by mathdanisur last updated on 13/Aug/21
$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:=\:? \\ $$
Commented by n0y0n last updated on 13/Aug/21
$$\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$$$ \\ $$$$\:\mathrm{But}\:\mathrm{you}\:\mathrm{can}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{right}\:\mathrm{limit} \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{1} \\ $$
Commented by mathdanisur last updated on 13/Aug/21
$$\mathrm{Thanks}\:\mathrm{ser},\:\mathrm{solution} \\ $$
Answered by puissant last updated on 13/Aug/21
$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{−} } \mathrm{x}^{\mathrm{x}} =\mathrm{1},\:\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \mathrm{x}^{\mathrm{x}} =\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{x}^{\mathrm{x}} \:=\:\mathrm{1}.. \\ $$
Commented by mathdanisur last updated on 14/Aug/21
$$\mathrm{Thanks}\:\mathrm{Ser} \\ $$