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