Question Number 119961 by bobhans last updated on 28/Oct/20
$${Without}\:{L}'{Hopital}\:{rule}\: \\ $$$$\:\underset{{x}\rightarrow\pi/\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)}{\mathrm{1}−\mathrm{2cos}\:{x}}\:? \\ $$
Answered by bramlexs22 last updated on 28/Oct/20
Answered by Dwaipayan Shikari last updated on 28/Oct/20
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{{sin}\left({x}−\frac{\pi}{\mathrm{3}}\right)}{\mathrm{1}−\mathrm{2}{cosx}}=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{sin}\left({x}−\frac{\pi}{\mathrm{3}}\right)}{{cos}\frac{\pi}{\mathrm{3}}−{cosx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{{x}−\frac{\pi}{\mathrm{3}}}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\right){sin}\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\right)}=\frac{\mathrm{2}}{\mathrm{4}}.\frac{{x}−\frac{\pi}{\mathrm{3}}}{{sin}\frac{\pi}{\mathrm{3}}.\left({x}−\frac{\pi}{\mathrm{3}}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$
Answered by bobhans last updated on 28/Oct/20
![lim_(x→π/3) ((sin (x−π/3))/((x−π/3))).(((x−π/3))/(1−2cos x)) = [ note lim_(x→π/3) ((sin (x−π/3))/(x−π/3)) = 1 ] lim_(x→π/3) ((x−π/3)/(1−2cos x)) = lim_(X→0) (X/(1−2cos (X+(π/3)))) lim_(X→0) (X/(1−2((1/2)cos X−((√3)/2)sin X)))= lim_(X→0) (X/(1−cos X+(√3)sin X)) = lim_(X→0) (1/(((1−cos X)/X)+(((√3)sin X)/X))) = lim_(X→0) (1/(((2sin^2 ((X/2)))/X) + (((√3) sin X)/X))) = (1/(0+(√3))) = (1/( (√3))).](https://www.tinkutara.com/question/Q119964.png)
$$\:\underset{{x}\rightarrow\pi/\mathrm{3}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left({x}−\pi/\mathrm{3}\right)}{\left({x}−\pi/\mathrm{3}\right)}.\frac{\left({x}−\pi/\mathrm{3}\right)}{\mathrm{1}−\mathrm{2cos}\:{x}}\:=\: \\ $$$$\left[\:{note}\:\underset{{x}\rightarrow\pi/\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}−\pi/\mathrm{3}\right)}{{x}−\pi/\mathrm{3}}\:=\:\mathrm{1}\:\right] \\ $$$$\underset{{x}\rightarrow\pi/\mathrm{3}} {\mathrm{lim}}\:\frac{{x}−\pi/\mathrm{3}}{\mathrm{1}−\mathrm{2cos}\:{x}}\:=\:\underset{{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{X}}{\mathrm{1}−\mathrm{2cos}\:\left({X}+\frac{\pi}{\mathrm{3}}\right)} \\ $$$$\underset{{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{X}}{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{X}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:{X}\right)}= \\ $$$$\underset{{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{X}}{\mathrm{1}−\mathrm{cos}\:{X}+\sqrt{\mathrm{3}}\mathrm{sin}\:{X}}\:=\:\underset{{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\frac{\mathrm{1}−\mathrm{cos}\:{X}}{{X}}+\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:{X}}{{X}}} \\ $$$$=\:\underset{{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{X}}{\mathrm{2}}\right)}{{X}}\:+\:\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{X}}{{X}}}\:=\:\frac{\mathrm{1}}{\mathrm{0}+\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}. \\ $$$$ \\ $$