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Question Number 114696 by bemath last updated on 20/Sep/20

lim_(x→0) ((sinh (2x)−sin 2x)/x^5 ) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sinh}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\mathrm{2}{x}}{{x}^{\mathrm{5}} }\:=? \\ $$

Answered by bobhans last updated on 20/Sep/20

e^(2x) =1+2x+2x^2 +((4x^3 )/3)+((2x^4 )/3)+((32x^5 )/(120))+... e^(2x) =1−2x+2x^2 −((4x^3 )/3)+((2x^4 )/3)−((32x^5 )/(120))+... sinh 2x=(1/2)(4x+((8x^3 )/3)+((64x^5 )/(120))+...) = 2x+((4x^3 )/3)+((32x^5 )/(120))+... lim_(x→0) ((sinh 2x−sin 2x)/x^5 ) = lim_(x→0) (((2x+((4x^3 )/3)+((32x^5 )/(120)))−(2x−((4x^3 )/3)+((32x^5 )/(120))))/x^5 )= lim_(x→0) (((8x^3 )/3)/x^5 ) = 0

$${e}^{\mathrm{2}{x}} =\mathrm{1}+\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}+... \\ $$$${e}^{\mathrm{2}{x}} =\mathrm{1}−\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{3}}−\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}+... \\ $$$$\mathrm{sinh}\:\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{x}+\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{64}{x}^{\mathrm{5}} }{\mathrm{120}}+...\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{x}+\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}+... \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sinh}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}}{{x}^{\mathrm{5}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{x}+\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}\right)−\left(\mathrm{2}{x}−\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}\right)}{{x}^{\mathrm{5}} }= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}}}{{x}^{\mathrm{5}} }\:=\:\mathrm{0}\: \\ $$

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