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Question Number 162643 by tounghoungko last updated on 31/Dec/21

lim_(x→∞) x^(4/3) (((x^2 +1))^(1/3) + ((3−x^2 ))^(1/3) ) =?

$$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}/\mathrm{3}} \:\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}^{\mathrm{2}} }\:\right)\:=? \\ $$

Answered by Ar Brandon last updated on 31/Dec/21

L=lim_(x→∞) x^(4/3) (((x^2 +1))^(1/3) +((3−x^2 ))^(1/3) ) =lim_(x→∞) x^(4/3) ∙x^(2/3) (1+(1/(3x^2 ))−(1−(3/(3x^2 )))) =lim_(x→∞) x^2 ((4/(3x^2 )))=(4/3)

$$\mathscr{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{4}/\mathrm{3}} \left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{4}/\mathrm{3}} \centerdot{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{3}{x}^{\mathrm{2}} }\right)\right) \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{3}{x}^{\mathrm{2}} }\right)=\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by Ar Brandon last updated on 31/Dec/21

Can someone please check. I have some doubt.

$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{please}\:\mathrm{check}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{some}\:\mathrm{doubt}. \\ $$

Commented by MJS_new last updated on 31/Dec/21

strange idea, not sure if it makes any sense (but anyway your result is right) x^(4/3) ((x^2 +1)^(1/3) −(x^2 −3)^(1/3) )=L a^(1/3) −b^(1/3) =c a−b−3a^(1/3) b^(1/3) (a^(1/3) −b^(1/3) )=c^3 3a^(1/3) b^(1/3) c=a−b−c^3 27abc^3 =(a−b−c^3 )^3 inserting & transforming x^(12) −((L^3 (54x^(10) −33x^8 −12L^3 x^4 +L^6 ))/(27L^3 −64))=0 ⇒ L≠(4/3) which means we cannot reach L=(3/4) while x∈R

$$\mathrm{strange}\:\mathrm{idea},\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{any}\:\mathrm{sense} \\ $$$$\left(\mathrm{but}\:\mathrm{anyway}\:\mathrm{your}\:\mathrm{result}\:\mathrm{is}\:\mathrm{right}\right) \\ $$$${x}^{\mathrm{4}/\mathrm{3}} \left(\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} −\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} \right)=\mathscr{L} \\ $$$${a}^{\mathrm{1}/\mathrm{3}} −{b}^{\mathrm{1}/\mathrm{3}} ={c} \\ $$$${a}−{b}−\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} \left({a}^{\mathrm{1}/\mathrm{3}} −{b}^{\mathrm{1}/\mathrm{3}} \right)={c}^{\mathrm{3}} \\ $$$$\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} {c}={a}−{b}−{c}^{\mathrm{3}} \\ $$$$\mathrm{27}{abc}^{\mathrm{3}} =\left({a}−{b}−{c}^{\mathrm{3}} \right)^{\mathrm{3}} \\ $$$$\mathrm{inserting}\:\&\:\mathrm{transforming} \\ $$$${x}^{\mathrm{12}} −\frac{\mathscr{L}^{\mathrm{3}} \left(\mathrm{54}{x}^{\mathrm{10}} −\mathrm{33}{x}^{\mathrm{8}} −\mathrm{12}\mathscr{L}^{\mathrm{3}} {x}^{\mathrm{4}} +\mathscr{L}^{\mathrm{6}} \right)}{\mathrm{27}\mathscr{L}^{\mathrm{3}} −\mathrm{64}}=\mathrm{0} \\ $$$$\Rightarrow\:\mathscr{L}\neq\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{which}\:\mathrm{means}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{reach}\:\mathscr{L}=\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{while}\:{x}\in\mathbb{R} \\ $$

Commented by Ar Brandon last updated on 31/Dec/21

Thanks for confirming, Sir.

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{confirming},\:\mathrm{Sir}. \\ $$

Answered by tounghoungko last updated on 31/Dec/21

lim_(x→∞) x^(4/3) (x^(2/3) ((1+(1/x^2 )))^(1/3) −x^(2/3) ((−(3/x^2 )+1))^(1/3) ) [ (1/x^2 ) = y ] = lim_(y→0) ((((1+y))^(1/3) −((1−3y))^(1/3) )/y) = lim_(y→0) (((1+(y/3))−(1−((3y)/3)))/y) = lim_(y→0) (((4/3)y)/y) = (4/3)

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}/\mathrm{3}} \:\left({x}^{\mathrm{2}/\mathrm{3}} \:\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:−{x}^{\mathrm{2}/\mathrm{3}} \:\sqrt[{\mathrm{3}}]{−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }+\mathrm{1}}\:\right) \\ $$$$\:\left[\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:{y}\:\right] \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+{y}}−\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{3}{y}}}{{y}}\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{{y}}{\mathrm{3}}\right)−\left(\mathrm{1}−\frac{\mathrm{3}{y}}{\mathrm{3}}\right)}{{y}} \\ $$$$=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{4}}{\mathrm{3}}{y}}{{y}}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by tounghoungko last updated on 31/Dec/21

lim_(x→∞) x^(4/3) (((x^2 +1))^(1/3) + ((3−x^2 ))^(1/3) ) = lim_(x→∞) x^(4/3 ) (((x^2 +1))^(1/3) +((−(x^2 −3)))^(1/3) ) = lim_(x→∞) x^(4/3) (((x^2 +1))^(1/3) −((x^2 −3))^(1/3) ) = lim_(x→0) x^2 (((1+(1/x^2 )))^(1/3) −((1−(3/x^2 )))^(1/3) ) = lim_(y→0) ((((1+y))^(1/3) −((1−3y))^(1/3) )/y) = lim_(y→0) (((1+(y/3))−(1−((3y)/3)))/y) = lim_(y→0) (((4/3)y)/y) = (4/3)

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}/\mathrm{3}} \:\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}^{\mathrm{2}} }\:\right) \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}/\mathrm{3}\:} \left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{−\left({x}^{\mathrm{2}} −\mathrm{3}\right)}\:\right) \\ $$$$\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}/\mathrm{3}} \:\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{1}}\:−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} −\mathrm{3}}\:\right) \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:−\sqrt[{\mathrm{3}}]{\mathrm{1}−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }}\:\right) \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+{y}}\:−\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{3}{y}}}{{y}} \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{{y}}{\mathrm{3}}\right)−\left(\mathrm{1}−\frac{\mathrm{3}{y}}{\mathrm{3}}\right)}{{y}} \\ $$$$\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{4}}{\mathrm{3}}{y}}{{y}}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Answered by mathmax by abdo last updated on 31/Dec/21

f(x)=x^(4/3) {(1+x^2 )^(1/3) −(^3 (√3))(1−(x^2 /3))^(1/3) } ⇒ f(x)=x^(4/3) {x^(2/3) (1+(1/x^2 ))^(1/3) −(^3 (√3))x^(2/3) ((1/x^2 )−(1/3))^(1/3) } =x^2 {(1+(1/x^2 ))^(1/3) −(1−(3/x^2 ))^(1/3) } ⇒f(x)∼x^2 (1+(1/(3x^2 ))−(1−(1/x^2 ))) =x^2 ((1/(3x^2 ))+(3/(3x^2 )))=(4/3) ⇒ lim_(x→+∞) f(x)=(4/3)

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} \left\{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right\}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} \left\{\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right)\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right\} \\ $$$$=\mathrm{x}^{\mathrm{2}} \left\{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right\} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3x}^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\right)\:=\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{3x}^{\mathrm{2}} }\right)=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4}}{\mathrm{3}} \\ $$

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