Question Number 195772 by dimentri last updated on 10/Aug/23
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Answered by MM42 last updated on 10/Aug/23
$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\frac{{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}\:−\mathrm{2}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}×{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{\mathrm{4}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\mathrm{3}{x}} \\ $$$$={lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\frac{\mathrm{1}\:−\mathrm{2}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{\mathrm{4}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\mathrm{3}{x}}\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\checkmark \\ $$$$= \\ $$
Answered by cortano12 last updated on 10/Aug/23
$$\:\:\mathrm{let}\:\frac{\mathrm{3x}}{\mathrm{2}}=\:\mathrm{u}\:\mathrm{then}\:\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}} \\ $$$$\:\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{u}−\mathrm{sin}\:\mathrm{2u}}{\mathrm{sin}\:\mathrm{4u}} \\ $$$$\:\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{u}\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{u}\right)}{\mathrm{2sin}\:\mathrm{2u}\:\mathrm{cos}\:\mathrm{2u}} \\ $$$$\:\:\mathrm{L}\:=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{u}\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{u}\right)}{\mathrm{4sin}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{2u}} \\ $$$$\:\:\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\mathrm{u}}{\mathrm{4sin}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{2u}} \\ $$$$\:\:\:=\:\frac{\mathrm{1}−\mathrm{2}}{\mathrm{4}\left(−\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$