Question-171235

Question Number 171235 by thean last updated on 10/Jun/22

Commented by eman_64 last updated on 10/Jun/22

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Commented by cortano1 last updated on 11/Jun/22

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

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

Answered by Mathspace last updated on 10/Jun/22

f(x)=(((√3)cosx−sinx)/(3(x−(π/3)))) we do the changement x−(π/3)=t (x→(π/3)⇔t→0) and f(x)=f(t+(π/3))=(((√3)cos(t+(π/3))−sin(t+(π/3)))/(3t)) =(((√3){(1/2)cost−((√3)/2)sint}−{(1/2)sint +((√3)/2)cost})/(3t)) =((((√3)/2)cost−(3/2)sint−(1/2)sint−((√3)/2)cost)/(3t)) =((−2sint)/(3t)) ⇒ lim_(t→0) ((−2sint)/(3t))=−(2/3)lim_(t→0) ((sint)/t) =−(2/3)×1=−(2/3) ⇒ lim_(x→(π/3)) f(x)=−(2/3) hospital method lim_(x→(π/3)) f(x)=lim_(x→(π/3)) ((−(√3)sinx−cosx)/3) =((−(√3).((√3)/2)−(1/2))/3)=((−(3/2)−(1/2))/3)=−(2/3)

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

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