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Question-206020




Question Number 206020 by NANIGOPAL last updated on 05/Apr/24
Answered by cortano12 last updated on 05/Apr/24
 = lim_(x→0)  ((x(((2tan x)/(1−tan^2 x)) −2tan x))/(4sin^4 x))   = lim_(x→0)  ((2xtan x(1−(1−tan^2 x)))/(4sin^4 x))    = (1/2) lim_(x→0)  ((tan^2 x)/(sin^2 x)) = (1/2)
$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\:−\mathrm{2tan}\:{x}\right)}{\mathrm{4sin}\:^{\mathrm{4}} {x}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}\mathrm{tan}\:{x}\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}\right)\right)}{\mathrm{4sin}\:^{\mathrm{4}} {x}} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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