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Question Number 126744 by bemath last updated on 24/Dec/20

  Given f(x)= ((2−(√3) cos x−sin x)/((6x−π)^2 ))   Find lim_(x→(π/6))  f(x) .

$$\:\:{Given}\:{f}\left({x}\right)=\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\left(\mathrm{6}{x}−\pi\right)^{\mathrm{2}} } \\ $$$$\:{Find}\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:{f}\left({x}\right)\:. \\ $$

Answered by liberty last updated on 24/Dec/20

 lim_(x→π/6) (((√3) sin x−cos x)/(12(6x−π))) = lim_(x→π/6) (((√3) cos x+sin x)/(72))   = (((√3) cos ((π/6))+sin ((π/6)))/(72)) = (((3/2)+(1/2))/(72)) = (1/(36))

$$\:\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{12}\left(\mathrm{6}{x}−\pi\right)}\:=\:\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{72}} \\ $$$$\:=\:\frac{\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}\right)+\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{72}}\:=\:\frac{\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{72}}\:=\:\frac{\mathrm{1}}{\mathrm{36}}\: \\ $$

Answered by bemath last updated on 24/Dec/20

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