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Question Number 164266 by naka3546 last updated on 15/Jan/22

lim_(x→∞) (x−2) − (√(x^2 +2x−5)) = ?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\left({x}−\mathrm{2}\right)\:−\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{5}}\:\:=\:\:? \\ $$

Answered by ajfour last updated on 15/Jan/22

x=(1/h)→∞ L=lim_(h→0) ((1−2h−(1+2h−5h^2 )^(1/2) )/h) =lim_(h→0) ((1−2h−(1+((2h−5h^2 )/2)))/h) = lim_(h→0) ((−4h−2h+5h^2 )/(2h)) =−3

$${x}=\frac{\mathrm{1}}{{h}}\rightarrow\infty \\ $$$${L}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2}{h}−\left(\mathrm{1}+\mathrm{2}{h}−\mathrm{5}{h}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} }{{h}} \\ $$$$\:\:\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2}{h}−\left(\mathrm{1}+\frac{\mathrm{2}{h}−\mathrm{5}{h}^{\mathrm{2}} }{\mathrm{2}}\right)}{{h}} \\ $$$$\:\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{4}{h}−\mathrm{2}{h}+\mathrm{5}{h}^{\mathrm{2}} }{\mathrm{2}{h}}\:=−\mathrm{3} \\ $$

Commented by naka3546 last updated on 15/Jan/22

Thank you, sir.

$${Thank}\:\:{you},\:\:{sir}. \\ $$

Answered by cortano1 last updated on 16/Jan/22

lim_(x→∞) (x−2)−(x+(2/(2.1))) = lim_(x→∞) x−2−x−1=−3

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}−\mathrm{2}\right)−\left({x}+\frac{\mathrm{2}}{\mathrm{2}.\mathrm{1}}\right)\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}−\mathrm{2}−{x}−\mathrm{1}=−\mathrm{3} \\ $$

Commented by naka3546 last updated on 16/Jan/22

thank you, sir.

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

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