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Question Number 117422 by bemath last updated on 11/Oct/20

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

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

Answered by bobhans last updated on 11/Oct/20

lim_(x→0) ((3tan 4x−4tan 3x)/(3sin 4x−4sin 3x)) = lim_(x→0) ((3(4x+(1/3)(4x)^3 )−4(3x+(1/3)(3x)^3 ))/(3(4x−(((4x)^3 )/6))−4(3x−(((3x)^3 )/6))))= lim_(x→0) ((12x+64x^3 −12x−36x^3 )/(12x−32x^3 −12x+18x^3 )) = lim_(x→0) ((28x^3 )/(−14x^3 )) = −2

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4x}−\mathrm{4tan}\:\mathrm{3x}}{\mathrm{3sin}\:\mathrm{4x}−\mathrm{4sin}\:\mathrm{3x}}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{4x}+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4x}\right)^{\mathrm{3}} \right)−\mathrm{4}\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3x}\right)^{\mathrm{3}} \right)}{\mathrm{3}\left(\mathrm{4x}−\frac{\left(\mathrm{4x}\right)^{\mathrm{3}} }{\mathrm{6}}\right)−\mathrm{4}\left(\mathrm{3x}−\frac{\left(\mathrm{3x}\right)^{\mathrm{3}} }{\mathrm{6}}\right)}= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{12x}+\mathrm{64x}^{\mathrm{3}} −\mathrm{12x}−\mathrm{36x}^{\mathrm{3}} }{\mathrm{12x}−\mathrm{32x}^{\mathrm{3}} −\mathrm{12x}+\mathrm{18x}^{\mathrm{3}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{28x}^{\mathrm{3}} }{−\mathrm{14x}^{\mathrm{3}} }\:=\:−\mathrm{2} \\ $$

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