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Question Number 128674 by bemath last updated on 09/Jan/21

lim_(x→0) ((3tan 4x−12tan x)/(3sin 4x−12sin x)) = ?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4x}−\mathrm{12tan}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{4x}−\mathrm{12sin}\:\mathrm{x}}\:=\:? \\ $$$$ \\ $$

Answered by liberty last updated on 09/Jan/21

Taylor series { ((tan x=x+(x^3 /3)+((2x^5 )/(15))+...)),((tan 4x=4x+((64x^3 )/3)+((2(4x)^5 )/(15))+...)) :} lim_(x→0) ((3(4x+((64x^3 )/3)+o(x^5 ))−12(x+(x^3 /3)+o(x^5 )))/(3(4x−((64x^3 )/6)+o(x^5 ))−12(x−(x^3 /6)+o(x^5 )))) = lim_(x→0) ((60x^3 −9(o(x^5 )))/(−30x^3 −9(o(x^5 )))) = −2.

$$\:\mathrm{Taylor}\:\mathrm{series}\:\begin{cases}{\mathrm{tan}\:\mathrm{x}=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2x}^{\mathrm{5}} }{\mathrm{15}}+...}\\{\mathrm{tan}\:\mathrm{4x}=\mathrm{4x}+\frac{\mathrm{64x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}\left(\mathrm{4x}\right)^{\mathrm{5}} }{\mathrm{15}}+...}\end{cases} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{4x}+\frac{\mathrm{64x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{5}} \right)\right)−\mathrm{12}\left(\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{5}} \right)\right)}{\mathrm{3}\left(\mathrm{4x}−\frac{\mathrm{64x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{5}} \right)\right)−\mathrm{12}\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{5}} \right)\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{60x}^{\mathrm{3}} −\mathrm{9}\left(\mathrm{o}\left(\mathrm{x}^{\mathrm{5}} \right)\right)}{−\mathrm{30x}^{\mathrm{3}} −\mathrm{9}\left(\mathrm{o}\left(\mathrm{x}^{\mathrm{5}} \right)\right)}\:=\:−\mathrm{2}. \\ $$

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