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lim-x-0-sin-tan-x-tan-sin-x-2x-cos-tan-x-2x-cos-sin-x-x-5-




Question Number 169966 by cortano1 last updated on 13/May/22
    lim_(x→0)  ((sin (tan x)−tan (sin x))/(2x cos (tan x)−2x cos (sin x)+x^5 )) =?
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)−\mathrm{tan}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{2}{x}\:\mathrm{cos}\:\left(\mathrm{tan}\:{x}\right)−\mathrm{2}{x}\:\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)+{x}^{\mathrm{5}} }\:=? \\ $$
Answered by qaz last updated on 13/May/22
sin tan x=x+(x^3 /6)−(x^5 /(40))−((55x^7 )/(1008))+...  tan sin x=x+(x^3 /6)−(x^5 /(40))−((107x^7 )/(5040))+...  2xcos tan x=2x−x^3 −((7x^5 )/(12))−((97x^7 )/(360))+...  2xcos sin x=2x−x^3 +((5x^5 )/(12))−((37x^7 )/(360))+...  ...
$$\mathrm{sin}\:\mathrm{tan}\:\mathrm{x}=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{40}}−\frac{\mathrm{55x}^{\mathrm{7}} }{\mathrm{1008}}+… \\ $$$$\mathrm{tan}\:\mathrm{sin}\:\mathrm{x}=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{40}}−\frac{\mathrm{107x}^{\mathrm{7}} }{\mathrm{5040}}+… \\ $$$$\mathrm{2xcos}\:\mathrm{tan}\:\mathrm{x}=\mathrm{2x}−\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{7x}^{\mathrm{5}} }{\mathrm{12}}−\frac{\mathrm{97x}^{\mathrm{7}} }{\mathrm{360}}+… \\ $$$$\mathrm{2xcos}\:\mathrm{sin}\:\mathrm{x}=\mathrm{2x}−\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{5x}^{\mathrm{5}} }{\mathrm{12}}−\frac{\mathrm{37x}^{\mathrm{7}} }{\mathrm{360}}+… \\ $$$$… \\ $$

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