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Question Number 54409 by pooja24 last updated on 03/Feb/19

lim_(x→∞) (√((x^2 +x+1)))−x=?  pls solve this

$${li}\underset{{x}\rightarrow\infty} {{m}}\sqrt{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−{x}=? \\ $$$${pls}\:{solve}\:{this} \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 03/Feb/19

we have (√(x^2  +x+1))=∣x∣(√(1+(1/x)+(1/x^2 )))∼1+(1/2)((1/x) +(1/x^2 )) (x→∞) ⇒  (√(x^2 +x+1))−x  ∼ ∣x∣{1+(1/(2x)) +(1/(2x^2 ))}−x ⇒  lim_(x→+∞) (√(x^2 +x+1)) −x =lim_(x→+∞) x{1+(1/(2x)) +(1/(2x^2 ))}−x  =lim_(x→+∞) ((1/2) +(1/(2x))) =(1/2)  lim_(x→−∞) (√(x^2 +x+1))−x =lim_(x→−∞) −x{1+(1/(2x)) +(1/(2x^2 ))}−x  =lim_(x→−∞)    −2x−(1/2) −(1/(2x)) =+∞ .

$${we}\:{have}\:\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}=\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\left({x}\rightarrow\infty\right)\:\Rightarrow \\ $$$$\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−{x}\:\:\sim\:\mid{x}\mid\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right\}−{x}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−{x}\:={lim}_{{x}\rightarrow+\infty} {x}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right\}−{x} \\ $$$$={lim}_{{x}\rightarrow+\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{x}\rightarrow−\infty} \sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−{x}\:={lim}_{{x}\rightarrow−\infty} −{x}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right\}−{x} \\ $$$$={lim}_{{x}\rightarrow−\infty} \:\:\:−\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}}\:=+\infty\:. \\ $$

Answered by kaivan.ahmadi last updated on 03/Feb/19

×(((√(x^2 +x+1))+x)/((√(x^2 +x+1))+1))=lim_(x→+∞) ((x+1)/((√(x^2 +x+1))+x))≈lim_(x→+∞) (x/(2x))=(1/2)    or    ≈lim_(x→∞) ∣x+(1/2)∣−x=(1/2)

$$×\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\mathrm{x}}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\mathrm{1}}=\mathrm{li}\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{m}}\frac{\mathrm{x}+\mathrm{1}}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\mathrm{x}}\approx\mathrm{li}\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{m}}\frac{\mathrm{x}}{\mathrm{2x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{or} \\ $$$$ \\ $$$$\approx\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\mid\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mid−\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Prithwish sen last updated on 03/Feb/19

let x=(1/h)  ∴x→∞⇒h→0  lim_(h→0) (((√(h^2 +h+1))−1)/h) (∵form (0/0) , we apply                                                    L′Hopital rule )  = lim_(h→0)  ((2h+1)/(2(√(h^2 +h+1))))  =(1/2)

$$\mathrm{let}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{h}} \\ $$$$\therefore\mathrm{x}\rightarrow\infty\Rightarrow\mathrm{h}\rightarrow\mathrm{0} \\ $$$$\mathrm{li}\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\sqrt{\mathrm{h}^{\mathrm{2}} +\mathrm{h}+\mathrm{1}}−\mathrm{1}}{\mathrm{h}}\:\left(\because\mathrm{form}\:\frac{\mathrm{0}}{\mathrm{0}}\:,\:\mathrm{we}\:\mathrm{apply}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule}\:\right) \\ $$$$=\:\mathrm{li}\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{m}}\:\frac{\mathrm{2h}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{h}^{\mathrm{2}} +\mathrm{h}+\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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