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Question Number 147685 by mathmax by abdo last updated on 22/Jul/21

calculate lim_(x→0)  ((sin(2sinx))/x^2 )

$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{sin}\left(\mathrm{2sinx}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$

Answered by liberty last updated on 23/Jul/21

lim_(x→0) ((sin (2sin x))/x^2 ) =  lim_(x→0) ((2sin x−((8sin^3 x)/6))/x^2 ) =  (1/6)lim_(x→0) ((12sin x−8sin^3 x)/x^2 ) =  (1/6)lim_(x→0) ((sin x(12−8sin^2 x))/x^2 ) =  (1/6)lim_(x→0) ((12−8sin^2 x)/x) = ∞

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{2sin}\:{x}\right)}{{x}^{\mathrm{2}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:{x}−\frac{\mathrm{8sin}\:^{\mathrm{3}} {x}}{\mathrm{6}}}{{x}^{\mathrm{2}} }\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{12sin}\:{x}−\mathrm{8sin}\:^{\mathrm{3}} {x}}{{x}^{\mathrm{2}} }\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}\left(\mathrm{12}−\mathrm{8sin}\:^{\mathrm{2}} {x}\right)}{{x}^{\mathrm{2}} }\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{12}−\mathrm{8sin}\:^{\mathrm{2}} {x}}{{x}}\:=\:\infty\: \\ $$

Answered by mathmax by abdo last updated on 23/Jul/21

sinx∼x−(x^3 /6) ⇒2sinx∼2x−(x^3 /3) ⇒sin(2sinx)∼sin(2x−(x^3 /3))  ∼2x−(x^3 /3)−(1/6)(2x−(x^3 /3))^3  ⇒((sin(2sinx))/x^2 )∼(2/x)−(x^2 /3)−(x/6)(2−(x^2 /3))^3  ⇒  lim_(x→0) ((sin(2sinx))/x^2 )=+^− ∞

$$\mathrm{sinx}\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{2sinx}\sim\mathrm{2x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\:\Rightarrow\mathrm{sin}\left(\mathrm{2sinx}\right)\sim\mathrm{sin}\left(\mathrm{2x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$$\sim\mathrm{2x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{2x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right)^{\mathrm{3}} \:\Rightarrow\frac{\mathrm{sin}\left(\mathrm{2sinx}\right)}{\mathrm{x}^{\mathrm{2}} }\sim\frac{\mathrm{2}}{\mathrm{x}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{x}}{\mathrm{6}}\left(\mathrm{2}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}\right)^{\mathrm{3}} \:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\mathrm{sin}\left(\mathrm{2sinx}\right)}{\mathrm{x}^{\mathrm{2}} }=\overset{−} {+}\infty \\ $$

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