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lim-x-0-3tan-4x-4tan-3x-3sin-4x-4sin-3x-




Question Number 91157 by jagoll last updated on 28/Apr/20
lim_(x→0)  ((3tan 4x−4tan 3x)/(3sin 4x−4sin 3x)) = ?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4}{x}−\mathrm{4tan}\:\mathrm{3}{x}}{\mathrm{3sin}\:\mathrm{4}{x}−\mathrm{4sin}\:\mathrm{3}{x}}\:=\:? \\ $$
Commented by john santu last updated on 28/Apr/20
lim_(x→0)  ((3(4x+(((4x)^3 )/3))−4(3x+(((3x)^3 )/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
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{4}{x}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}} }{\mathrm{3}}\right)−\mathrm{4}\left(\mathrm{3}{x}+\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{3}}\right)}{\mathrm{3}\left(\mathrm{4}{x}−\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}} }{\mathrm{6}}\right)−\mathrm{4}\left(\mathrm{3}{x}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{6}}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{12}{x}+\mathrm{64}{x}^{\mathrm{3}} \right)−\left(\mathrm{12}{x}+\mathrm{36}{x}^{\mathrm{3}} \right)}{\left(\mathrm{12}{x}−\mathrm{32}{x}^{\mathrm{3}} \right)−\left(\mathrm{12}{x}−\mathrm{18}{x}^{\mathrm{3}} \right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{28}{x}^{\mathrm{3}} }{−\mathrm{14}{x}^{\mathrm{3}} }\:=\:−\mathrm{2}\: \\ $$

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