Question Number 117412 by bobhans last updated on 11/Oct/20
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\mathrm{x}}{\:\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{x}}\:=\:? \\ $$
Commented by Lordose last updated on 11/Oct/20
$$\mathrm{you}\:\mathrm{edited}\:\mathrm{the}\:\mathrm{question}.? \\ $$
Commented by bobhans last updated on 11/Oct/20
$$\mathrm{yes}.\:\mathrm{typo} \\ $$
Answered by Lordose last updated on 11/Oct/20
$$\frac{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}\:=\:\frac{\mathrm{0}}{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}}\:=\:\mathrm{0} \\ $$
Answered by Olaf last updated on 11/Oct/20
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{−\mathrm{2cos}{x}}{−\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right)}\:=\:\frac{−\mathrm{2}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{−\sqrt{\mathrm{3}}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left(\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}\right) \\ $$
Answered by bemath last updated on 11/Oct/20
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\mathrm{x}}{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{x}}\:=? \\ $$$$\mathrm{Solution}: \\ $$$$\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\mathrm{letting}\:\mathrm{w}\:=\:\mathrm{x}−\frac{\pi}{\mathrm{6}}\: \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\left(\mathrm{w}+\frac{\pi}{\mathrm{6}}\right)}{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\mathrm{w}+\frac{\pi}{\mathrm{6}}\right)}\:=\: \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{w}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{w}\right)}{\mathrm{1}−\sqrt{\mathrm{3}}\left(\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\mathrm{tan}\:\mathrm{w}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}\:\mathrm{w}}\right)}\:= \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{w}−\mathrm{cos}\:\mathrm{w}}{\mathrm{1}−\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{w}}{\:\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{w}}\right)}\:= \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{w}\right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{w}−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{w}\right)}{\:\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{w}−\sqrt{\mathrm{3}}−\mathrm{3}\:\mathrm{tan}\:\mathrm{w}}\:= \\ $$$$\sqrt{\mathrm{3}}\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{w}}{−\mathrm{4tan}\:\mathrm{w}}\:= \\ $$$$\sqrt{\mathrm{3}}\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)\:\left(\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)−\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)\right)}{−\mathrm{4tan}\:\mathrm{w}}= \\ $$$$\frac{\sqrt{\mathrm{3}}}{−\mathrm{4}}\:×\left(−\sqrt{\mathrm{3}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by TANMAY PANACEA last updated on 11/Oct/20
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}\:\frac{}{}} {\mathrm{li}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{{sin}\frac{\pi}{\mathrm{6}}−{sinx}}{{tan}\frac{\pi}{\mathrm{6}}−{tanx}}\right) \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2}{cos}\left(\frac{\frac{\pi}{\mathrm{6}}+{x}}{\mathrm{2}}\right){sin}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)}{\frac{{sin}\left(\frac{\pi}{\mathrm{6}}−{x}\right)}{{cos}\left(\frac{\pi}{\mathrm{6}}\right){cosx}}} \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\mathrm{2}{cos}\frac{\pi}{\mathrm{6}}{cosx}×{cos}\left(\frac{\frac{\pi}{\mathrm{6}}+{x}}{\mathrm{2}}\right){sin}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)×\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)×{coz}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$