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Question Number 148328 by liberty last updated on 27/Jul/21

lim_(x→0) ((sin 2x+2sin^2 x−2sin x)/(cos x−cos^2 x)) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}+\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=? \\ $$

Answered by EDWIN88 last updated on 27/Jul/21

Answered by mathmax by abdo last updated on 27/Jul/21

let f(x)=((sin(2x)+2sin^2 (x)−2sinx)/(cosx−cos^2 x)) we have sin(2x)∼2x ,sin^2 x∼x^2 cosx−cos^2 x=cosx(1−cosx)∼(1−(x^2 /2)).(x^2 /2)=(x^2 /2)−(x^4 /4) ⇒ f(x)∼((2x+2x^2 −2x)/((x^2 /2)−(x^4 /4)))=(2/((1/2)−(x^2 /4))) ⇒lim_(x→0) f(x)=4

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sin}\left(\mathrm{2x}\right)+\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{2sinx}}{\mathrm{cosx}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{sin}\left(\mathrm{2x}\right)\sim\mathrm{2x}\:\:\:\:\:\:\:\:,\mathrm{sin}^{\mathrm{2}} \mathrm{x}\sim\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{cosx}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}=\mathrm{cosx}\left(\mathrm{1}−\mathrm{cosx}\right)\sim\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right).\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{2x}+\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}}{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}}=\frac{\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{4} \\ $$

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