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Question Number 90210 by manuel__ last updated on 22/Apr/20

lim_(x→(π/3)) (((1−2cos (x))/(π−3x)))=?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−\mathrm{2cos}\:\left({x}\right)}{\pi−\mathrm{3}{x}}\right)=? \\ $$

Commented by mathmax by abdo last updated on 22/Apr/20

f(x)=((2cosx−1)/(3x−π)) =((2cosx−1)/(3(x−(π/3)))) we do the changement x−(π/3)=t ⇒ (x→(π/3)⇔ t→0) and f(x)=((2cos(t+(π/3))−1)/(3t))=g(t) =((2(cost cos((π/3))−sint sin((π/3)))−1)/(3t)) =((cost−(√3)sint −1)/(3t)) =−((1−cost)/(3t)) −((√3)/3) ((sint)/t) 1−cost ∼(t^2 /2) and ((sint)/t) ∼ 1 ⇒g(t) ∼−(1/(3t))×(t^2 /2)=−(t/6)→0 ⇒lim_(t→0) g(t) =−((√3)/3) =lim_(x→(π/3)) f(x)

$${f}\left({x}\right)=\frac{\mathrm{2}{cosx}−\mathrm{1}}{\mathrm{3}{x}−\pi}\:=\frac{\mathrm{2}{cosx}−\mathrm{1}}{\mathrm{3}\left({x}−\frac{\pi}{\mathrm{3}}\right)}\:{we}\:{do}\:{the}\:{changement}\:{x}−\frac{\pi}{\mathrm{3}}={t}\:\Rightarrow \\ $$$$\left({x}\rightarrow\frac{\pi}{\mathrm{3}}\Leftrightarrow\:{t}\rightarrow\mathrm{0}\right)\:{and}\:{f}\left({x}\right)=\frac{\mathrm{2}{cos}\left({t}+\frac{\pi}{\mathrm{3}}\right)−\mathrm{1}}{\mathrm{3}{t}}={g}\left({t}\right) \\ $$$$=\frac{\mathrm{2}\left({cost}\:{cos}\left(\frac{\pi}{\mathrm{3}}\right)−{sint}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)\right)−\mathrm{1}}{\mathrm{3}{t}} \\ $$$$=\frac{{cost}−\sqrt{\mathrm{3}}{sint}\:−\mathrm{1}}{\mathrm{3}{t}}\:=−\frac{\mathrm{1}−{cost}}{\mathrm{3}{t}}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\frac{{sint}}{{t}} \\ $$$$\mathrm{1}−{cost}\:\sim\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\:\:{and}\:\frac{{sint}}{{t}}\:\sim\:\mathrm{1}\:\:\Rightarrow{g}\left({t}\right)\:\sim−\frac{\mathrm{1}}{\mathrm{3}{t}}×\frac{{t}^{\mathrm{2}} }{\mathrm{2}}=−\frac{{t}}{\mathrm{6}}\rightarrow\mathrm{0} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:\:{g}\left({t}\right)\:=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:={lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:{f}\left({x}\right) \\ $$

Answered by john santu last updated on 22/Apr/20

L′hopital lim_(x→(π/3)) ((2sin x)/(−3)) = −(2/3)×sin ((π/3)) = −(2/3)×((√3)/2) = −((√3)/3)

$${L}'{hopital}\: \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{x}}{−\mathrm{3}}\:=\:−\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$=\:−\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

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