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lim-x-0-sin-38x-38sin-x-19x-3-




Question Number 84932 by jagoll last updated on 17/Mar/20
lim_(x→0)  ((sin 38x−38sin x)/(19x^3 )) =
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{38x}−\mathrm{38sin}\:\mathrm{x}}{\mathrm{19x}^{\mathrm{3}} }\:=\: \\ $$
Commented by john santu last updated on 17/Mar/20
lim_(x→0)  ((38cos 38x − 38cos x)/(3.19x^2 )) =  lim_(x→0)  ((2cos 38x−cos x)/(3x^2 )) =   lim_(x→0)  ((−76sin 38x+sin x)/(6x)) =  ((−76×38+1)/6) = −((2887)/6)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{38cos}\:\mathrm{38x}\:−\:\mathrm{38cos}\:\mathrm{x}}{\mathrm{3}.\mathrm{19x}^{\mathrm{2}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:\mathrm{38x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{3x}^{\mathrm{2}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{76sin}\:\mathrm{38x}+\mathrm{sin}\:\mathrm{x}}{\mathrm{6x}}\:= \\ $$$$\frac{−\mathrm{76}×\mathrm{38}+\mathrm{1}}{\mathrm{6}}\:=\:−\frac{\mathrm{2887}}{\mathrm{6}} \\ $$

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