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

Question Number 120686 by bobhans last updated on 01/Nov/20

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

Answered by john santu last updated on 02/Nov/20

standart way lim_(x→0) ((sin x−tan x)/x^3 ) = lim_(x→0) ((tan x(cos x−1))/x^3 ) = lim_(x→0) ((tan x(−2sin^2 ((1/2)x)))/x^3 ) = −0.5

$${standart}\:{way}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}−\mathrm{tan}\:{x}}{{x}^{\mathrm{3}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)}{{x}^{\mathrm{3}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}\left(−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)\right)}{{x}^{\mathrm{3}} }\:=\:−\mathrm{0}.\mathrm{5} \\ $$

Answered by mathmax by abdo last updated on 02/Nov/20

let f(x)=((sinx−tanx)/x^3 ) ⇒f(x) =((sinx−((sinx)/(cosx)))/x^3 )=((sinxcosx−sinx)/(x^3 cosx)) =((sinx)/x).((cosx−1)/(x^2 cosx)) we have ((sinx)/x)∼1 and cosx−1∼−(x^2 /2) ⇒ ((cosx)/(x^2 cosx)) ∼−(1/2) ⇒lim_(x→0) f(x)=−(1/2)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}−\mathrm{tanx}}{\mathrm{x}^{\mathrm{3}} }\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sinx}−\frac{\mathrm{sinx}}{\mathrm{cosx}}}{\mathrm{x}^{\mathrm{3}} }=\frac{\mathrm{sinxcosx}−\mathrm{sinx}}{\mathrm{x}^{\mathrm{3}} \:\mathrm{cosx}} \\ $$$$=\frac{\mathrm{sinx}}{\mathrm{x}}.\frac{\mathrm{cosx}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{cosx}}\:\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{sinx}}{\mathrm{x}}\sim\mathrm{1}\:\mathrm{and}\:\mathrm{cosx}−\mathrm{1}\sim−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} \mathrm{cosx}}\:\sim−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by malwaan last updated on 02/Nov/20

lim_(x→0) (((x− (x^3 /(3!)) + ..)−(x + (x^3 /3) +..))/x^3 ) =lim_(x→0) (− (1/6) − (1/3))= − (1/2)

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left({x}−\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\:..\right)−\left({x}\:+\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+..\right)}{{x}^{\mathrm{3}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(−\:\frac{\mathrm{1}}{\mathrm{6}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\right)=\:−\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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