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Question Number 55592 by gunawan last updated on 27/Feb/19

lim_(x→π/3) ((cos x−sin (π/6))/((π/6)−(x/2)))=..

$$\underset{{x}\rightarrow\pi/\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}}{\frac{\pi}{\mathrm{6}}−\frac{{x}}{\mathrm{2}}}=.. \\ $$

Commented by maxmathsup by imad last updated on 27/Feb/19

let A(x)=((cosx−sin((π/6)))/((π/6)−(x/2))) ⇒ A(x)=6((cosx−(1/2))/(π−3x)) =3 ((2cosx−1)/(3((π/3)−x))) =((1−2cosx)/(x−(π/3))) changement x−(π/3) =t give lim_(x→(π/3)) A(x) =lim_(t→0) ((1−2cos(t+(π/3)))/t) =lim_(t→0) ((1−2{(1/2)cost −((√3)/2)sint)})/t) =lim_(t→0) ((1−cost +(√3)sint)/t) =lim_(t→0) ((1−cost)/t) +lim_(t→0) (√3)((sint)/t) but cost ∼1−(t^2 /2) ⇒((1−cost)/t) ∼ (t/2) →0(t→0) also sint ∼t ⇒lim_(t→0) ((sint)/t) =1 ⇒ lim_(x→(π/3)) A(x)=(√3).

$${let}\:{A}\left({x}\right)=\frac{{cosx}−{sin}\left(\frac{\pi}{\mathrm{6}}\right)}{\frac{\pi}{\mathrm{6}}−\frac{{x}}{\mathrm{2}}}\:\Rightarrow\:{A}\left({x}\right)=\mathrm{6}\frac{{cosx}−\frac{\mathrm{1}}{\mathrm{2}}}{\pi−\mathrm{3}{x}}\:=\mathrm{3}\:\:\frac{\mathrm{2}{cosx}−\mathrm{1}}{\mathrm{3}\left(\frac{\pi}{\mathrm{3}}−{x}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{2}{cosx}}{{x}−\frac{\pi}{\mathrm{3}}}\:\:\:{changement}\:{x}−\frac{\pi}{\mathrm{3}}\:={t}\:{give}\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {A}\left({x}\right) \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\mathrm{1}−\mathrm{2}{cos}\left({t}+\frac{\pi}{\mathrm{3}}\right)}{{t}}\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{\left.\mathrm{1}−\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{2}}{cost}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sint}\right)\right\}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cost}\:+\sqrt{\mathrm{3}}{sint}}{{t}}\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cost}}{{t}}\:+{lim}_{{t}\rightarrow\mathrm{0}} \:\:\sqrt{\mathrm{3}}\frac{{sint}}{{t}} \\ $$$${but}\:{cost}\:\sim\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}−{cost}}{{t}}\:\sim\:\frac{{t}}{\mathrm{2}}\:\rightarrow\mathrm{0}\left({t}\rightarrow\mathrm{0}\right)\:\:{also}\:{sint}\:\sim{t}\:\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:\frac{{sint}}{{t}}\:=\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\:{A}\left({x}\right)=\sqrt{\mathrm{3}}. \\ $$

Answered by kaivan.ahmadi last updated on 27/Feb/19

hopital lim_(x→(π/3)) ((−sinx)/(−(1/2)))=lim_(x→(π/3)) 2sinx=2sin(π/3)=2×((√3)/2)=(√3)

$${hopital} \\ $$$${li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {{m}}\frac{−{sinx}}{−\frac{\mathrm{1}}{\mathrm{2}}}={li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {{m}}\mathrm{2}{sinx}=\mathrm{2}{sin}\frac{\pi}{\mathrm{3}}=\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}} \\ $$

Answered by math1967 last updated on 27/Feb/19

lim_(x→(π/3)) ((cosx−cos((π/2)−(π/6)))/((π/6)−(x/2))) lim_(x→(π/3)) ((2sin((π/6)−(x/2))sin((x/2)+(π/6)))/((π/6)−(x/2))) ★ 2×1×sin((π/6)+(π/6))=2×1×((√3)/2)=(√3) ★lim_(((π/6)−(x/2))→0) ((sin((π/6)−(x/2)))/(((π/6)−(x/2))))=1 ∵x→(π/3) ∴(x/2)→(π/6) ∴((π/6)−(x/2))→0

$$\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{\mathrm{cosx}−\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\right)}{\frac{\pi}{\mathrm{6}}−\frac{\mathrm{x}}{\mathrm{2}}}\: \\ $$$$\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\left(\frac{\pi}{\mathrm{6}}−\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\right)}{\frac{\pi}{\mathrm{6}}−\frac{\mathrm{x}}{\mathrm{2}}}\:\bigstar \\ $$$$\mathrm{2}×\mathrm{1}×\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{6}}\right)=\mathrm{2}×\mathrm{1}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}}\: \\ $$$$\bigstar\underset{\left(\frac{\pi}{\mathrm{6}}−\frac{\mathrm{x}}{\mathrm{2}}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}−\frac{\mathrm{x}}{\mathrm{2}}\right)}{\left(\frac{\pi}{\mathrm{6}}−\frac{\mathrm{x}}{\mathrm{2}}\right)}=\mathrm{1} \\ $$$$\because\mathrm{x}\rightarrow\frac{\pi}{\mathrm{3}}\:\:\therefore\frac{\mathrm{x}}{\mathrm{2}}\rightarrow\frac{\pi}{\mathrm{6}}\:\therefore\left(\frac{\pi}{\mathrm{6}}−\frac{\mathrm{x}}{\mathrm{2}}\right)\rightarrow\mathrm{0} \\ $$$$ \\ $$

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