Question-118040

Question Number 118040 by mathocean1 last updated on 14/Oct/20

Answered by Ar Brandon last updated on 14/Oct/20

lim_(x→(π/3)) ((1−2cosx)/(π−3x))=(0/0)=lim_(x→(π/3)) ((2sinx)/(−3))=2×((√3)/2)÷−3=−(1/( (√3)))

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2cos}{x}}{\pi−\mathrm{3}{x}}=\frac{\mathrm{0}}{\mathrm{0}}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{\mathrm{2sin}{x}}{−\mathrm{3}}=\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{\div}−\mathrm{3}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$

Answered by bobhans last updated on 15/Oct/20

L=lim_(x→π/3) ((1−2cos x)/(π−3x)) set x=(π/3)+u , u→0 L=lim_(u→0) ((1−2cos (u+(π/3)))/(π−(π+3u))) L=lim_(u→0) ((1−2((1/2)cos u−(1/2)(√3) sin u))/(−3u)) L=lim_(u→0) ((1−cos u+(√3) sin u)/(−3u)) L=lim_(u→0) ((2sin^2 ((u/2))+2(√3) sin ((u/2))cos ((u/2)))/(−3u)) L=lim_(u→0) ((2sin ((u/2)){sin ((u/2))+(√3) cos ((u/2))})/(−3u)) L= −(1/3)(0+(√3)) = −((√3)/3)

$$\mathrm{L}=\underset{{x}\rightarrow\pi/\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2cos}\:\mathrm{x}}{\pi−\mathrm{3x}} \\ $$$$\mathrm{set}\:\mathrm{x}=\frac{\pi}{\mathrm{3}}+\mathrm{u}\:,\:\mathrm{u}\rightarrow\mathrm{0} \\ $$$$\mathrm{L}=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2cos}\:\left(\mathrm{u}+\frac{\pi}{\mathrm{3}}\right)}{\pi−\left(\pi+\mathrm{3u}\right)} \\ $$$$\mathrm{L}=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{u}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{u}\right)}{−\mathrm{3u}} \\ $$$$\mathrm{L}=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{u}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{u}}{−\mathrm{3u}} \\ $$$$\mathrm{L}=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{u}}{\mathrm{2}}\right)+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{u}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{u}}{\mathrm{2}}\right)}{−\mathrm{3u}} \\ $$$$\mathrm{L}=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\left(\frac{\mathrm{u}}{\mathrm{2}}\right)\left\{\mathrm{sin}\:\left(\frac{\mathrm{u}}{\mathrm{2}}\right)+\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\frac{\mathrm{u}}{\mathrm{2}}\right)\right\}}{−\mathrm{3u}} \\ $$$$\mathrm{L}=\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{0}+\sqrt{\mathrm{3}}\right)\:=\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

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