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Question Number 146549 by mathmax by abdo last updated on 13/Jul/21
find lim_(x→0) ((cos(x−sinx)+1−cos(x^2 ))/x^2 )
$$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{cos}\left(\mathrm{x}−\mathrm{sinx}\right)+\mathrm{1}−\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} } \\ $$
Answered by Olaf_Thorendsen last updated on 14/Jul/21
((cos(x−sinx)+1−cos(x^2 ))/x^2 ) ∼_0 ((cos(x−x+(x^3 /6))+1−(1−(x^4 /2)))/x^2 ) ∼_0 (((1−(1/2)((x^3 /6))^2 )+(x^4 /2))/x^2 ) ∼_0 (((1−(1/2)((x^3 /6))^2 )+(x^4 /2))/x^2 ) →_0 ∞
$$\frac{\mathrm{cos}\left({x}−\mathrm{sin}{x}\right)+\mathrm{1}−\mathrm{cos}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$$\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{cos}\left({x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)+\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)}{{x}^{\mathrm{2}} } \\ $$$$\underset{\mathrm{0}} {\sim}\:\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)^{\mathrm{2}} \right)+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} } \\ $$$$\underset{\mathrm{0}} {\sim}\:\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)^{\mathrm{2}} \right)+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} }\:\underset{\mathrm{0}} {\rightarrow}\:\infty \\ $$
Answered by mathmax by abdo last updated on 15/Jul/21
let f(x)=((cos(x−sinx)+1−cos(x^2 ))/x^2 ) we have sinx∼x−(x^3 /6) ⇒x−sinx∼(x^3 /6) ⇒cos(x−sinx)∼cos((x^3 /6)) ∼1−(x^6 /(72)) and 1−cos(x^2 )∼(x^4 /2) ⇒f(x)∼((1−(x^6 /(72))+(x^4 /2))/x^2 )=(1/x^2 )−(x^4 /(72))+(x^2 /2)→+∞ (x→0) ⇒lim_(x→0) f(x)=+∞
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{cos}\left(\mathrm{x}−\mathrm{sinx}\right)+\mathrm{1}−\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{x}−\mathrm{sinx}\sim\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{cos}\left(\mathrm{x}−\mathrm{sinx}\right)\sim\mathrm{cos}\left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right) \\ $$$$\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{72}}\:\:\mathrm{and}\:\mathrm{1}−\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\sim\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{72}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{72}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\rightarrow+\infty \\ $$$$\left(\mathrm{x}\rightarrow\mathrm{0}\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=+\infty \\ $$

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