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Question Number 184796 by Spillover last updated on 11/Jan/23

Evaluate lim_(x→(π/6)) (((√3)sin x−cos x)/(x−(π/6)))

$${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{{x}−\frac{\pi}{\mathrm{6}}} \\ $$

Commented by MJS_new last updated on 11/Jan/23

2

$$\mathrm{2} \\ $$

Commented by MJS_new last updated on 11/Jan/23

these are all easy with l′Ho^� pital

$$\mathrm{these}\:\mathrm{are}\:\mathrm{all}\:\mathrm{easy}\:\mathrm{with}\:\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital} \\ $$

Answered by aba last updated on 12/Jan/23

let t=x−(π/6) ⇒ x=t+(π/6) lim_(x→(π/6)) (((√3)sin(x)−cos(x))/(x−(π/6)))=lim_(t→0 ) (((√3)sin(t+(π/6))−cos(t+(π/6)))/t) =lim_(t→0) (((√3)(sin(t)cos((π/6))+cos(t)sin((π/6)))−(cos(t)cos((π/6))−sin(t)sin((π/6))))/t) =lim_(t→0) (((√3)((√3)sin(t)+cos(t))−((√3)cos(t)−sin(t)))/(2t)) =lim_(t→0) ((3sin(t)+sin(t)+(√3)sin(t)−(√3)cos(t))/(2t)) =lim_(t→0) ((4sin(t))/(2t)) =2

$$\mathrm{let}\:\mathrm{t}=\mathrm{x}−\frac{\pi}{\mathrm{6}}\:\Rightarrow\:\mathrm{x}=\mathrm{t}+\frac{\pi}{\mathrm{6}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\mathrm{sin}\left(\mathrm{x}\right)−\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{x}−\frac{\pi}{\mathrm{6}}}=\underset{\mathrm{t}\rightarrow\mathrm{0}\:} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\mathrm{sin}\left(\mathrm{t}+\frac{\pi}{\mathrm{6}}\right)−\mathrm{cos}\left(\mathrm{t}+\frac{\pi}{\mathrm{6}}\right)}{\mathrm{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\left(\mathrm{sin}\left(\mathrm{t}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)+\mathrm{cos}\left(\mathrm{t}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)−\left(\mathrm{cos}\left(\mathrm{t}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)−\mathrm{sin}\left(\mathrm{t}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)}{\mathrm{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}\mathrm{sin}\left(\mathrm{t}\right)+\mathrm{cos}\left(\mathrm{t}\right)\right)−\left(\sqrt{\mathrm{3}}\mathrm{cos}\left(\mathrm{t}\right)−\mathrm{sin}\left(\mathrm{t}\right)\right)}{\mathrm{2t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3sin}\left(\mathrm{t}\right)+\mathrm{sin}\left(\mathrm{t}\right)+\sqrt{\mathrm{3}}\mathrm{sin}\left(\mathrm{t}\right)−\sqrt{\mathrm{3}}\mathrm{cos}\left(\mathrm{t}\right)}{\mathrm{2t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4sin}\left(\mathrm{t}\right)}{\mathrm{2t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2} \\ $$

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