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Question Number 87378 by jagoll last updated on 04/Apr/20

lim_(x→0) ((tan 3x−3tan x)/x^3 )

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{3x}−\mathrm{3tan}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} } \\ $$

Commented by john santu last updated on 04/Apr/20

lim_(x→0) (((3x+(1/3)(3x)^3 +o((3x)^3 ))−3(x+(1/3)x^3 +o(x^3 )))/x^3 ) = lim_(x→0) ((9x^3 −x^3 )/x^3 ) = 8

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3x}\right)^{\mathrm{3}} +\mathrm{o}\left(\left(\mathrm{3x}\right)^{\mathrm{3}} \right)\right)−\mathrm{3}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)\right)}{\mathrm{x}^{\mathrm{3}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{9x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{3}} }\:=\:\mathrm{8} \\ $$

Answered by ajfour last updated on 04/Apr/20

L=lim_(x→0) (((((3tan x−tan^3 x)/(1−3tan^2 x)))−3tan x)/x^3 ) =lim_(x→0) ((tan x)/x)×lim_(x→0) (((((3−tan^2 x−3+9tan^2 x)/(1−3tan^2 x))))/x^2 ) = 1×8 = 8 .

$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{3tan}\:{x}−\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}}\right)−\mathrm{3tan}\:{x}}{{x}^{\mathrm{3}} } \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:{x}}{{x}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} {x}−\mathrm{3}+\mathrm{9tan}\:^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}}\right)}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\mathrm{1}×\mathrm{8}\:=\:\mathrm{8}\:. \\ $$

Commented by peter frank last updated on 04/Apr/20

thank you

$${thank}\:{you} \\ $$

Commented by jagoll last updated on 04/Apr/20

lim_(x→0) ((3tan x−tan^3 x −3tan x+9tan^3 x)/x^3 ) lim_(x→0) ((8tan^3 x)/x^3 ) = 8 ← the ans

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{x}−\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\:−\mathrm{3tan}\:\mathrm{x}+\mathrm{9tan}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8tan}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:=\:\mathrm{8}\:\leftarrow\:\mathrm{the}\:\mathrm{ans} \\ $$

Commented by jagoll last updated on 04/Apr/20

haha..same sir. typo

$$\mathrm{haha}..\mathrm{same}\:\mathrm{sir}.\:\mathrm{typo} \\ $$

Commented by ajfour last updated on 04/Apr/20

nevr mind, both of us, wouldn′t have scored in an exam!

$${nevr}\:{mind},\:{both}\:{of}\:{us},\:{wouldn}'{t} \\ $$$${have}\:{scored}\:{in}\:{an}\:{exam}! \\ $$

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