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Question Number 91158 by jagoll last updated on 28/Apr/20

lim_(x→∞) (√(4x^2 +2x)) − ((8x^3 +4x^2 ))^(1/(3 ))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}}\:−\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} } \\ $$

Commented by jagoll last updated on 28/Apr/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by john santu last updated on 28/Apr/20

lim_(x→∞) 2x(√(1+(1/(2x)))) −2x ((1+(1/(2x))))^(1/(3 )) = [ let (1/(2x)) = p ⇒p→0 ] lim_(p→0) (((√(1+p))−((1+p))^(1/(3 )) )/p) = lim_(p→0) (((1+(p/2))−(1+(p/3)))/p) = lim_(p→0) ((((p/6)))/p) = (1/6)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}}\:−\mathrm{2}{x}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}}\:= \\ $$$$\left[\:{let}\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:=\:{p}\:\Rightarrow{p}\rightarrow\mathrm{0}\:\right] \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{p}}−\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}+{p}}}{{p}}\:=\:\: \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{{p}}{\mathrm{2}}\right)−\left(\mathrm{1}+\frac{{p}}{\mathrm{3}}\right)}{{p}}\:=\: \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{{p}}{\mathrm{6}}\right)}{{p}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by abdomathmax last updated on 28/Apr/20

let f(x)=(√(4x^2 +2x))−^3 (√(8x^3 +4x^2 )) we have (√(4x^2 +2x))=2x(√(1+(1/(2x))))∼2x(1+(1/(4x))) (x→+∞) (8x^3 +4x^2 )^(1/3) =2x(1+(1/(2x)))^(1/3) ∼2x(1+(1/(6x))) ⇒ f(x)∼2x +(1/2)−2x−(1/3) =(1/6) ⇒lim_(x→+∞) f(x)=(1/6)

$${let}\:{f}\left({x}\right)=\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{2}{x}}−^{\mathrm{3}} \sqrt{\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} } \\ $$$${we}\:{have}\:\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{2}{x}}=\mathrm{2}{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}}\sim\mathrm{2}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}}\right)\:\left({x}\rightarrow+\infty\right) \\ $$$$\left(\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\sim\mathrm{2}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}{x}}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\mathrm{2}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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