Question Number 169315 by Mastermind last updated on 28/Apr/22
$${lim}_{{x}\rightarrow\infty} \left(\frac{\mathrm{1}+\sqrt{{x}+\mathrm{2}}}{\mathrm{1}−\sqrt{{x}+\mathrm{2}}}\right) \\ $$$$ \\ $$$${Mastermind} \\ $$
Commented by infinityaction last updated on 28/Apr/22
$$\:\:\:\:\:\:\:{p}\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\cancel{{x}}}{\cancel{{x}}}\left(\frac{\mathrm{1}/{x}\:+\sqrt{\mathrm{1}+\mathrm{2}/{x}}}{\mathrm{1}/{x}−\sqrt{\mathrm{1}+\mathrm{2}/{x}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{p}\:=\:\left(\frac{\mathrm{0}+\sqrt{\mathrm{1}+\mathrm{0}}}{\mathrm{0}−\sqrt{\mathrm{1}+\mathrm{0}}}\right) \\ $$$$=\:−\mathrm{1} \\ $$$$ \\ $$
Answered by alephzero last updated on 28/Apr/22
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\sqrt{{x}+\mathrm{2}}}{\mathrm{1}−\sqrt{{x}+\mathrm{2}}}\:= \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}−\sqrt{{x}+\mathrm{2}}}+\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{x}+\mathrm{2}}}{\mathrm{1}−\sqrt{{x}+\mathrm{2}}}\:= \\ $$$$=\:\mathrm{0}+\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{x}+\mathrm{2}}}{\mathrm{1}−\sqrt{{x}+\mathrm{2}}} \\ $$$$\sqrt{{x}+\mathrm{2}}\:\sim\:\sqrt{{x}+\mathrm{2}}−\mathrm{1}\:{as}\:{x}\rightarrow\infty \\ $$$$\Rightarrow\:\sqrt{{x}+\mathrm{2}}\:\sim\:−\left(\mathrm{1}−\sqrt{{x}+\mathrm{2}}\right) \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{x}+\mathrm{2}}}{\mathrm{1}−\sqrt{{x}+\mathrm{2}}}\:=\:−\mathrm{1} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{x}+\mathrm{2}}}{\mathrm{1}−\sqrt{{x}+\mathrm{2}}}\:=\:−\mathrm{1} \\ $$