lim-x-x-2-x-3-x-x-3-1-1-7-cos-1-x-

Question Number 141463 by bramlexs22 last updated on 19/May/21

$$\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \left(\:\sqrt[{\mathrm{7}\:\:}]{\frac{{x}^{\mathrm{3}} +{x}}{{x}^{\mathrm{3}} +\mathrm{1}}}\:−\mathrm{cos}\:\frac{\mathrm{1}}{{x}}\right)? \\ $$

Commented by jcarlos last updated on 19/May/21

Commented by Adetunji last updated on 19/May/21

Answered by mathmax by abdo last updated on 19/May/21

f(x)=x^2 {(((x^3 +x)/(x^3 +1)))^(1/7) −cos((1/x))} we have (((x^3 +x)/(x^3 +1)))^(1/7) =(1+((x−1)/(x^3 +1)))^(1/7) =(1+((x(1−(1/x)))/(x^3 (1+(1/x^3 )))))^(1/7) =(1+(1/x^2 ).((1−(1/x))/(1+x^(−3) )))^(1/7) ∼(1+(1/x^2 )(1−(1/x))(1−(1/x^3 )))^(1/7) =(1+(1/x^2 )(1−(1/x^3 )−(1/x)+(1/x^4 )))^(1/7) =(1+(1/x^2 )−(1/x^5 )−(1/x^3 )+(1/x^6 ))^(1/7) ∼1+(1/7)((1/x^2 )−(1/x^5 )−(1/x^3 )+(1/x^6 )) ∼1+(1/(7x^2 )) and cos((1/x))∼1−(1/(2x^2 )) +(1/(4!x^4 )) ⇒x^2 {.....} ∼x^2 +(1/7)−(1/(7x^3 ))−(1/(7x))+(1/(7x^4 ))−x^2 +(1/2)−(1/(4!x^2 )) ⇒lim_(x→+∞) f(x) =(1/7)+(1/2) =(9/(14))

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \left\{\left(\frac{\mathrm{x}^{\mathrm{3}} \:+\mathrm{x}}{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} −\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right\}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left(\frac{\mathrm{x}^{\mathrm{3}} \:+\mathrm{x}}{\mathrm{x}^{\mathrm{3}} \:+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \:=\left(\mathrm{1}+\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \:+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} =\left(\mathrm{1}+\frac{\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }.\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{1}+\mathrm{x}^{−\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \:\sim\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right)\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \:=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{6}} }\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{6}} }\right)\:\:\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7x}^{\mathrm{2}} }\:\:\mathrm{and}\:\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\sim\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}!\mathrm{x}^{\mathrm{4}} } \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \left\{…..\right\}\:\sim\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{7x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{7x}}+\frac{\mathrm{1}}{\mathrm{7x}^{\mathrm{4}} }−\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}!\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{9}}{\mathrm{14}} \\ $$$$ \\ $$

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