Menu Close

lim-x-0-1-cos-1-cos-x-x-x-x-x-

Tinku Tara 0 Comments
Pin




Question Number 32407 by saru53424@gmail.com last updated on 24/Mar/18
lim_(x→0) ((1−cos (1−cos x))/(x×x×x×x))
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}×\boldsymbol{\mathrm{x}}×\boldsymbol{\mathrm{x}}×\boldsymbol{\mathrm{x}}} \\ $$
Commented by abdo imad last updated on 24/Mar/18
if you mean lim_(x→0) ((1−cos(1−cosx))/x^4 ) in this case its better to use hospital theorem let put u(x)=1−cos(1−cosx) and v(x)=x^4 ⇒ v^′ (x)=4x^3 ⇒v^(′′) (x)=12x^2 ⇒v^((3)) (x)=24x ⇒v^((4)) (x)=24 u^′ (x) =sinx sin(1−cosx)⇒u^(′′) (x)=cosx sin(1−cosx) +sinx.sinx cos(1−cosx)=cosxsin(1−cosx) +sin^2 x cos(1−cosx) u^((3)) (x)=−sinxsin(1−cosx) +cosx.sinx cos(1−cosx) +2sinx cosx.cos(1−cosx) −sin^2 x sin(1−cosx) =−sinxsin(1−cosx) +(1/2)sin(2x)cos(1−cosx) +sin(2x).cos(1−cosx)−sin^2 xsin(1−cosx) u^((4)) (x)=−cosx sin(1−cosx)−sinx sinxcos(1−cosx) +cos(2x)cos(1−cosx)−(1/2)sin(2x)sinx sin(1−cosx) +2cos(2x)cos(1−cosx)−sin(2x)sinx sin(1−cosx) −2sinx cosx sin(1−cosx)−sin^3 cos(1−cosx) u^((4)) (0)=1 +2 =3 ⇒lim_(x→0) ((1−cos(1−cosx))/x^4 ) =(3/(24)) =(1/8) +(perhaps).....
$${if}\:{you}\:{mean}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cosx}\right)}{{x}^{\mathrm{4}} }\:{in}\:{this}\:{case}\:{its} \\ $$$${better}\:{to}\:{use}\:{hospital}\:{theorem}\:{let}\:{put}\:{u}\left({x}\right)=\mathrm{1}−{cos}\left(\mathrm{1}−{cosx}\right) \\ $$$${and}\:{v}\left({x}\right)={x}^{\mathrm{4}} \Rightarrow\:{v}^{'} \left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} \:\Rightarrow{v}^{''} \left({x}\right)=\mathrm{12}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{v}^{\left(\mathrm{3}\right)} \left({x}\right)=\mathrm{24}{x}\:\Rightarrow{v}^{\left(\mathrm{4}\right)} \left({x}\right)=\mathrm{24} \\ $$$${u}^{'} \left({x}\right)\:={sinx}\:{sin}\left(\mathrm{1}−{cosx}\right)\Rightarrow{u}^{''} \left({x}\right)={cosx}\:{sin}\left(\mathrm{1}−{cosx}\right) \\ $$$$+{sinx}.{sinx}\:{cos}\left(\mathrm{1}−{cosx}\right)={cosxsin}\left(\mathrm{1}−{cosx}\right) \\ $$$$+{sin}^{\mathrm{2}} {x}\:{cos}\left(\mathrm{1}−{cosx}\right) \\ $$$${u}^{\left(\mathrm{3}\right)} \left({x}\right)=−{sinxsin}\left(\mathrm{1}−{cosx}\right)\:+{cosx}.{sinx}\:{cos}\left(\mathrm{1}−{cosx}\right) \\ $$$$+\mathrm{2}{sinx}\:{cosx}.{cos}\left(\mathrm{1}−{cosx}\right)\:−{sin}^{\mathrm{2}} {x}\:{sin}\left(\mathrm{1}−{cosx}\right) \\ $$$$=−{sinxsin}\left(\mathrm{1}−{cosx}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{1}−{cosx}\right) \\ $$$$+{sin}\left(\mathrm{2}{x}\right).{cos}\left(\mathrm{1}−{cosx}\right)−{sin}^{\mathrm{2}} {xsin}\left(\mathrm{1}−{cosx}\right) \\ $$$${u}^{\left(\mathrm{4}\right)} \left({x}\right)=−{cosx}\:{sin}\left(\mathrm{1}−{cosx}\right)−{sinx}\:{sinxcos}\left(\mathrm{1}−{cosx}\right) \\ $$$$+{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{1}−{cosx}\right)−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right){sinx}\:{sin}\left(\mathrm{1}−{cosx}\right) \\ $$$$+\mathrm{2}{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{1}−{cosx}\right)−{sin}\left(\mathrm{2}{x}\right){sinx}\:{sin}\left(\mathrm{1}−{cosx}\right) \\ $$$$−\mathrm{2}{sinx}\:{cosx}\:{sin}\left(\mathrm{1}−{cosx}\right)−{sin}^{\mathrm{3}} {cos}\left(\mathrm{1}−{cosx}\right) \\ $$$${u}^{\left(\mathrm{4}\right)} \left(\mathrm{0}\right)=\mathrm{1}\:+\mathrm{2}\:=\mathrm{3}\:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cosx}\right)}{{x}^{\mathrm{4}} }\:=\frac{\mathrm{3}}{\mathrm{24}}\:=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$+\left({perhaps}\right)….. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *