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lim-x-pi-2-sin2x-tgx-




Question Number 145835 by mathdanisur last updated on 08/Jul/21
lim_(x→(π/2)) (sin2x∙tgx)=?
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{lim}}\left({sin}\mathrm{2}{x}\centerdot{tgx}\right)=? \\ $$
Answered by bramlexs22 last updated on 09/Jul/21
 lim_(x→(π/2))  ((2sin x cos x.sin x)/(cos x)) = lim_(x→(π/2)) 2sin^2 x  = 2×1 = 2
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}.\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\:=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}2sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$=\:\mathrm{2}×\mathrm{1}\:=\:\mathrm{2} \\ $$
Answered by puissant last updated on 08/Jul/21
lim_(x→(π/2)) ((2sin(x)cos(x)sin(x))/(cos(x)))  =lim_(x→(π/2)) 2sin^2 (x)=lim_(x→(π/2)) 1−cos(2x)  =1−(−1)=2...
$$\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right) \\ $$$$=\mathrm{1}−\left(−\mathrm{1}\right)=\mathrm{2}… \\ $$

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