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lim-0-sin-cos-sin-




Question Number 121813 by bemath last updated on 12/Nov/20
  lim_(θ→0)  ((sin θ−θ cos θ)/(sin θ−θ)) =?
$$\:\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\theta−\theta\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta−\theta}\:=? \\ $$
Answered by bobhans last updated on 12/Nov/20
  lim_(θ→0)  ((sin θ−θ cos θ)/(sin θ−θ)) =    lim_(θ→0)  ((θ−(θ^3 /6)−θ(1−(θ^2 /2)))/(θ−(θ^3 /6)−θ)) =   lim_(θ→0)  (((((2θ^3 )/6)))/(−((θ^3 /6)))) = −2.
$$\:\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\theta−\theta\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta−\theta}\:=\: \\ $$$$\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\theta−\frac{\theta^{\mathrm{3}} }{\mathrm{6}}−\theta\left(\mathrm{1}−\frac{\theta^{\mathrm{2}} }{\mathrm{2}}\right)}{\theta−\frac{\theta^{\mathrm{3}} }{\mathrm{6}}−\theta}\:= \\ $$$$\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{2}\theta^{\mathrm{3}} }{\mathrm{6}}\right)}{−\left(\frac{\theta^{\mathrm{3}} }{\mathrm{6}}\right)}\:=\:−\mathrm{2}. \\ $$
Answered by liberty last updated on 12/Nov/20
 lim_(θ→0)  ((cos θ−(cos θ−θ sin θ))/(cos θ−1)) =   lim_(θ→0)  ((θ sin θ)/(−2sin^2 (θ/2))) = (1/(−2((1/4)))) = −2.▲
$$\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\theta−\left(\mathrm{cos}\:\theta−\theta\:\mathrm{sin}\:\theta\right)}{\mathrm{cos}\:\theta−\mathrm{1}}\:= \\ $$$$\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\theta\:\mathrm{sin}\:\theta}{−\mathrm{2sin}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right)}\:=\:\frac{\mathrm{1}}{−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\:−\mathrm{2}.\blacktriangle \\ $$
Answered by Dwaipayan Shikari last updated on 12/Nov/20
lim_(x→0) ((x−(x^3 /(3!))−x+(x^3 /(2!)))/(x−(x^3 /6)−x))=lim_(x→0) ((x^3 /3)/(−(x^3 /6)))=−2
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}!}}{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}}{−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}=−\mathrm{2} \\ $$

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