Question Number 155625 by puissant last updated on 02/Oct/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\:? \\ $$
Answered by yeti123 last updated on 03/Oct/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:{x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}\:+\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+\:…}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}\:+\:\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:×\:\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\:×\:\frac{\mathrm{3}}{{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}\:+\:\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:×\:\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{e}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$
Commented by puissant last updated on 03/Oct/21
$${thanks}.. \\ $$
Answered by cortano last updated on 03/Oct/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} =\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\right)} \\ $$$$=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$