Question Number 31246 by malwaan last updated on 04/Mar/18
$$\mathrm{without}\:\mathrm{using}\:\mathrm{lohpital} \\ $$$$\mathrm{find} \\ $$$$\underset{\mathrm{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sinx}}{\mathrm{cos}\:\mathrm{3x}} \\ $$
Commented by malwaan last updated on 04/Mar/18
$$\mathrm{please} \\ $$$$\mathrm{I}\:\mathrm{need}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{very}\:\mathrm{soon} \\ $$
Commented by prof Abdo imad last updated on 04/Mar/18
$${ch}\:{x}=\frac{\pi}{\mathrm{6}}−{t}\:{give} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{6}}} \:\frac{\mathrm{1}−\mathrm{2}{sinx}}{{cos}\left(\mathrm{3}{x}\right)}={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{6}}−{t}\right)}{{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{3}{t}\right)} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}{cost}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sint}\right)}{{sin}\left(\mathrm{3}{t}\right)} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cost}\:+\sqrt{\mathrm{3}}\:{sint}}{{sin}\left(\mathrm{3}{t}\right)}\:{but} \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cost}}{{sin}\left(\mathrm{3}{t}\right)}\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\frac{\mathrm{1}−{cost}}{{t}}}{\frac{\mathrm{3}\:{sin}\left(\mathrm{3}{t}\right)}{\mathrm{3}{t}}}=\frac{\mathrm{0}}{\mathrm{3}}=\mathrm{0}\:{and} \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\sqrt{\mathrm{3}}\:{sint}}{{sin}\left(\mathrm{3}{t}\right)}={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\:\:\sqrt{\mathrm{3}}\:\:\frac{\frac{{sint}}{{t}}}{\frac{\mathrm{3}{sin}\left(\mathrm{3}{t}\right)}{\mathrm{3}{t}}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:. \\ $$
Answered by naka3546 last updated on 04/Mar/18
Commented by malwaan last updated on 05/Mar/18
$$\mathrm{thank}\:\mathrm{you} \\ $$