Question Number 82541 by zainal tanjung last updated on 22/Feb/20
$$\:\underset{\mathrm{x}\rightarrow\infty} {\mathrm{Lim}}\:\sqrt{\mathrm{4x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{3}}\:−\mathrm{2x}−\mathrm{4} \\ $$
Commented by john santu last updated on 22/Feb/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}\:−\left(\mathrm{2}{x}+\mathrm{4}\right)\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}\:−\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{16}}\:= \\ $$$$\frac{−\mathrm{8}−\mathrm{16}}{\mathrm{2}.\mathrm{2}}\:=\:−\mathrm{6}\: \\ $$
Commented by mathmax by abdo last updated on 22/Feb/20
$${lim}_{{x}\rightarrow−\infty} \sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}−\mathrm{2}{x}−\mathrm{4}\:=+\infty \\ $$$${at}\:+\infty\:\:{let}\:{f}\left({x}\right)=\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}−\mathrm{2}{x}−\mathrm{4} \\ $$$${we}\:{have}\:{f}\left({x}\right)=\mathrm{2}{x}\sqrt{\mathrm{1}−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} }}−\mathrm{2}{x}−\mathrm{4} \\ $$$$\sim\mathrm{2}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} }\right)\right)−\mathrm{2}{x}−\mathrm{4}\:=\mathrm{2}{x}−\mathrm{2}+\frac{\mathrm{3}}{\mathrm{4}{x}}−\mathrm{2}{x}−\mathrm{4}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{3}}{\mathrm{4}{x}}\:−\mathrm{6}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=−\mathrm{6} \\ $$