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Question Number 99676 by bemath last updated on 22/Jun/20

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

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

Answered by bobhans last updated on 23/Jun/20

lim_(x→0) ((sin x−((sin^3 x)/6) −x)/(x(1−((sin^2 x)/2) −1))) = lim_(x→0) ((x−(x^3 /6) −((sin^3 )/6) −x)/(x (−((sin^2 x)/2))))  = lim_(x→0) ((−(x^3 /3))/(−(x^3 /2))) = (1/3) × (2/1) = (2/3) ■

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

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