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Question Number 83668 by jagoll last updated on 05/Mar/20

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

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

Answered by john santu last updated on 05/Mar/20

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

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

Commented by jagoll last updated on 05/Mar/20

thank you mister

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

Commented by jagoll last updated on 05/Mar/20

Dalil L′Hopital lim_(x→0) ((cos 2x−1)/(2x sin (2x)+2x^2 cos (2x)))= lim_(x→0) ((−2sin^2 (x))/(2x sin (2x) +2x^2 cos (2x))) = lim_(x→0) ((−2sin^2 x)/(2x (sin (2x)+x cos (2x)))) = ((−2)/(2(2+1))) = −(1/3)

$$\mathrm{Dalil}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\mathrm{cos}\:\mathrm{2x}−\mathrm{1}}{\mathrm{2x}\:\mathrm{sin}\:\left(\mathrm{2x}\right)+\mathrm{2x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\mathrm{2x}\right)}= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:^{\mathrm{2}} \:\left(\mathrm{x}\right)}{\mathrm{2x}\:\mathrm{sin}\:\left(\mathrm{2x}\right)\:+\mathrm{2x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\mathrm{2x}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{2x}\:\left(\mathrm{sin}\:\left(\mathrm{2x}\right)+\mathrm{x}\:\mathrm{cos}\:\left(\mathrm{2x}\right)\right)}\:= \\ $$$$\frac{−\mathrm{2}}{\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)}\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$

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