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Question Number 182829 by yaslm last updated on 14/Dec/22
Answered by cortano1 last updated on 15/Dec/22
L= lim_(x→(π/6)) ((2sin x−1)/( (√3) sec x−2)) = lim_(x→(π/6)) ((cos x(2sin x−1))/( (√3)−2cos x)) = (1/2)(√3) ×lim_(x→(π/6)) ((2sin x−1)/( (√3)−2cos x)) [ let x=(π/6)+y ] L=(1/2)(√3) lim_(y→0) ((2sin (y+(π/6))−1)/( (√3)−2cos (y+(π/6)))) = (1/2)(√3) ×lim_(y→0) ((2((1/( 2 ))(√3) sin y+(1/2)cos y)−1)/( (√3)−2((1/2)(√3) cos y−(1/2)sin y))) =(1/2)(√3) ×lim_(y→0) (((√3) sin y+cos y−1)/( (√3)−(√3)cos y+sin y)) =(1/2)(√3) ×lim_(y→0) (((√3)sin y−2sin^2 ((1/2)y))/( 2(√3) sin^2 ((1/2)y)+sin y)) =(1/2)(√3) ×lim_(y→0) ((2(√3) cos ((1/2)y)−2sin ((1/2)y))/(2(√3)sin ((1/2)y)+2cos ((1/2)y))) = (1/2)(√3) ×((2(√3))/2) = (3/2)
$$\:{L}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}\:\mathrm{sec}\:{x}−\mathrm{2}} \\ $$$$\:=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}\left(\mathrm{2sin}\:{x}−\mathrm{1}\right)}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}\:{x}} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}\:{x}} \\ $$$$\left[\:{let}\:{x}=\frac{\pi}{\mathrm{6}}+{y}\:\right] \\ $$$${L}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\left({y}+\frac{\pi}{\mathrm{6}}\right)−\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}\:\left({y}+\frac{\pi}{\mathrm{6}}\right)} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\:\mathrm{2}\:}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{y}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{y}\right)−\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{cos}\:{y}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{y}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{y}+\mathrm{cos}\:{y}−\mathrm{1}}{\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\mathrm{cos}\:{y}+\mathrm{sin}\:{y}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:{y}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)}{\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)+\mathrm{sin}\:{y}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)−\mathrm{2sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)}{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)+\mathrm{2cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 15/Dec/22
L=lim_(x→(π/6)) ((2sinx−1)/( (√3)secx−2))=lim_(x→(π/6)) ((2sinxcosx−cosx)/( (√3)−2cosx)) =lim_(x→(π/6)) ((sin2x−cosx)/( (√3)−2cosx))=lim_(t→0) ((sin((π/3)−2t)−cos((π/6)−t))/( (√3)−2cos((π/6)−t))) =lim_(t→0) ((((√3)/2)cos2t−(1/2)sin2t−((√3)/2)cost−(1/2)sint)/( (√3)−(√3)cost−sint)) =lim_(t→0) ((((√3)/2)(1−2t^2 )−t−((√3)/2)(1−(t^2 /2))−(t/2))/( (√3)−(√3)(1−(t^2 /2))−t)) =lim_(t→0) ((−(√3)t^2 −((3t)/2)+(((√3)t^2 )/4))/((t^2 /2)−t))=(3/2)
$$\mathscr{L}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{\mathrm{2sin}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}\mathrm{sec}{x}−\mathrm{2}}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{\mathrm{2sin}{x}\mathrm{cos}{x}−\mathrm{cos}{x}}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}{x}} \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{\mathrm{sin2}{x}−\mathrm{cos}{x}}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}{x}}=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}−\mathrm{2}{t}\right)−\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}−{t}\right)}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}\left(\frac{\pi}{\mathrm{6}}−{t}\right)} \\ $$$$\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos2}{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{t}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{t}}{\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\mathrm{cos}{t}−\mathrm{sin}{t}} \\ $$$$\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)−{t}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)−\frac{{t}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)−{t}} \\ $$$$\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\sqrt{\mathrm{3}}{t}^{\mathrm{2}} −\frac{\mathrm{3}{t}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}{t}^{\mathrm{2}} }{\mathrm{4}}}{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−{t}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by malwan last updated on 15/Dec/22
lim_(x→(π/6)) ((2(sin x − (1/2)))/(((√3)/(cos x)) −2)) =lim_(x→(π/6)) ((2cosx(sinx−sin(π/6)))/(2(cos(π/6)−cosx))) =lim_(x→(π/6)) ((((√3)/2)(2cos((x/2)+(π/(12)))sin((x/2)−(π/(12))))/(−2sin((π/(12))+x)sin((π/(12))−(x/(12))))) =((((√3)/2)×((√3)/2))/(1/2)) = (3/2)
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {{lim}}\:\frac{\mathrm{2}\left({sin}\:{x}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\right)}{\frac{\sqrt{\mathrm{3}}}{{cos}\:{x}}\:−\mathrm{2}} \\ $$$$=\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {{lim}}\:\frac{\mathrm{2}{cosx}\left({sinx}−{sin}\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}\left({cos}\frac{\pi}{\mathrm{6}}−{cosx}\right)} \\ $$$$=\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {{lim}}\:\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{2}{cos}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{12}}\right){sin}\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{12}}\right)\right.}{−\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{12}}+{x}\right){sin}\left(\frac{\pi}{\mathrm{12}}−\frac{{x}}{\mathrm{12}}\right)} \\ $$$$=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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