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Question Number 117412 by bobhans last updated on 11/Oct/20

lim_(x→(π/6)) ((1−2sin x)/( 1−(√3) tan x)) = ?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\mathrm{x}}{\:\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{x}}\:=\:? \\ $$

Commented by Lordose last updated on 11/Oct/20

you edited the question.?

$$\mathrm{you}\:\mathrm{edited}\:\mathrm{the}\:\mathrm{question}.? \\ $$

Commented by bobhans last updated on 11/Oct/20

yes. typo

$$\mathrm{yes}.\:\mathrm{typo} \\ $$

Answered by Lordose last updated on 11/Oct/20

((1−2((1/2)))/( (√3)−((√3)/3))) = (0/((2(√3))/3)) = 0

$$\frac{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}\:=\:\frac{\mathrm{0}}{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}}\:=\:\mathrm{0} \\ $$

Answered by Olaf last updated on 11/Oct/20

lim_(x→(π/6)) ((−2cosx)/(−(√3)(1+tan^2 x))) = ((−2((√3)/2))/(−(√3)(1+((1/( (√3))))^2 ))) = (3/4) (Hospital′s rule)

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{−\mathrm{2cos}{x}}{−\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right)}\:=\:\frac{−\mathrm{2}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{−\sqrt{\mathrm{3}}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left(\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}\right) \\ $$

Answered by bemath last updated on 11/Oct/20

lim_(x→(π/6)) ((1−2sin x)/(1−(√3) tan x)) =? Solution: Without L′Hopital letting w = x−(π/6) lim_(w→0) ((1−2sin (w+(π/6)))/(1−(√3) tan (w+(π/6)))) = lim_(w→0) ((1−2(((√3)/2) sin w+(1/2)cos w))/(1−(√3)((((1/( (√3)))+tan w)/(1−(1/( (√3)))tan w))))) = lim_(w→0) ((1−(√3) sin w−cos w)/(1−(√3)(((1+(√3) tan w)/( (√3)−tan w))))) = lim_(w→0) ((((√3)−tan w)(1−cos w−(√3) sin w))/( (√3)−tan w−(√3)−3 tan w)) = (√3) lim_(w→0) ((2sin^2 ((1/2)w)−(√3) sin w)/(−4tan w)) = (√3) lim_(w→0) ((2sin ((1/2)w) (sin ((1/2)w)−(√3) cos ((1/2)w)))/(−4tan w))= ((√3)/(−4)) ×(−(√3)) = (3/4)

$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\mathrm{x}}{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{x}}\:=? \\ $$$$\mathrm{Solution}: \\ $$$$\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\mathrm{letting}\:\mathrm{w}\:=\:\mathrm{x}−\frac{\pi}{\mathrm{6}}\: \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\left(\mathrm{w}+\frac{\pi}{\mathrm{6}}\right)}{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\mathrm{w}+\frac{\pi}{\mathrm{6}}\right)}\:=\: \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{w}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{w}\right)}{\mathrm{1}−\sqrt{\mathrm{3}}\left(\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\mathrm{tan}\:\mathrm{w}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}\:\mathrm{w}}\right)}\:= \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{w}−\mathrm{cos}\:\mathrm{w}}{\mathrm{1}−\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{w}}{\:\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{w}}\right)}\:= \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{w}\right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{w}−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{w}\right)}{\:\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{w}−\sqrt{\mathrm{3}}−\mathrm{3}\:\mathrm{tan}\:\mathrm{w}}\:= \\ $$$$\sqrt{\mathrm{3}}\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{w}}{−\mathrm{4tan}\:\mathrm{w}}\:= \\ $$$$\sqrt{\mathrm{3}}\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)\:\left(\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)−\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}\right)\right)}{−\mathrm{4tan}\:\mathrm{w}}= \\ $$$$\frac{\sqrt{\mathrm{3}}}{−\mathrm{4}}\:×\left(−\sqrt{\mathrm{3}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by TANMAY PANACEA last updated on 11/Oct/20

li_(x→(π/6) (/)) lim_(x→(π/6)) (2/( (√3)))(((sin(π/6)−sinx)/(tan(π/6)−tanx))) (2/( (√3)))lim_(x→(π/6)) ((2cos((((π/6)+x)/2))sin((((π/6)−x)/2)))/((sin((π/6)−x))/(cos((π/6))cosx))) (2/( (√3)))lim_(x→(π/6)) 2cos(π/6)cosx×cos((((π/6)+x)/2))sin((((π/6)−x)/2))×(1/(2sin((((π/6)−x)/2))×coz((((π/6)−x)/2)))) =(2/( (√3)))×((√3)/2)×((√3)/2)×((√3)/2)×(1/1)=(3/4)

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}\:\frac{}{}} {\mathrm{li}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{{sin}\frac{\pi}{\mathrm{6}}−{sinx}}{{tan}\frac{\pi}{\mathrm{6}}−{tanx}}\right) \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2}{cos}\left(\frac{\frac{\pi}{\mathrm{6}}+{x}}{\mathrm{2}}\right){sin}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)}{\frac{{sin}\left(\frac{\pi}{\mathrm{6}}−{x}\right)}{{cos}\left(\frac{\pi}{\mathrm{6}}\right){cosx}}} \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\mathrm{2}{cos}\frac{\pi}{\mathrm{6}}{cosx}×{cos}\left(\frac{\frac{\pi}{\mathrm{6}}+{x}}{\mathrm{2}}\right){sin}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)×\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)×{coz}\left(\frac{\frac{\pi}{\mathrm{6}}−{x}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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