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Question Number 145205 by imjagoll last updated on 03/Jul/21

lim_(x→0) ((1−cos 2(sin (sin x)))/x^2 )=?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\left(\mathrm{sin}\:\left(\mathrm{sin}\:\mathrm{x}\right)\right)}{\mathrm{x}^{\mathrm{2}} }=? \\ $$

Answered by EDWIN88 last updated on 03/Jul/21

lim_(x→0) ((1−cos 2(sin (sin x)))/x^2 ) = lim_(x→0) ((2sin^2 (sin (sin x)))/x^2 ) =2lim_(x→0) ((sin^2 (sin (sin x)))/((sin (sin x))^2 )) .lim_(x→0) (((sin (sin x))^2 )/x^2 ) = 2×1×lim_(x→0) (((sin (sin x))/(sin (x))))^2 ×lim_(x→0) ((sin^2 (x))/x^2 ) =2×1×1×1=2

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right)}{{x}^{\mathrm{2}} }\: \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right)}{{x}^{\mathrm{2}} }\: \\ $$$$=\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right)}{\left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right)^{\mathrm{2}} }\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}×\mathrm{1}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:\left({x}\right)}\right)^{\mathrm{2}} ×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{2}×\mathrm{1}×\mathrm{1}×\mathrm{1}=\mathrm{2} \\ $$

Answered by mathmax by abdo last updated on 03/Jul/21

f(x)=((1−cos(2(sin(sinx)))/x^2 ) we have sinx∼x ⇒sin(sinx)∼x ⇒cos(2sin(sinx))∼cos(2x)∼1−(((2x)^2 )/2) =1−2x^2 ⇒ 1−cos(2sin(sinx))∼2x^2 ⇒f(x)∼((2x^2 )/x^2 ) =2 ⇒lim_(x→0) f(x)=2

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

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