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lim-x-0-tanx-x-1-x-2-




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

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