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Question Number 128605 by john_santu last updated on 08/Jan/21

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

$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} }=? \\ $$

Commented by john_santu last updated on 08/Jan/21

Answered by mathmax by abdo last updated on 09/Jan/21

f(x)=(√(x^2 +x+1))−(√(x^2 −x+1)) ⇒ f(x)=∣x∣(√(1+(1/x)+(1/x^2 )))−∣x∣(√(1−(1/x) +(1/x^2 ))) for x→−∞ f(x)∼−x(1+(1/2)((1/x)+(1/x^2 )))+x(1+(1/2)(−(1/x)+(1/x^2 )) =−(1/2)−(1/(2x))−(1/2) +(1/(2x))=−1 ⇒lim_(x→−∞) f(x)=−1

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

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