Question Number 131370 by EDWIN88 last updated on 10/Feb/21
$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}}\:=? \\ $$
Answered by JDamian last updated on 04/Feb/21
$$\infty \\ $$
Answered by EDWIN88 last updated on 10/Feb/21
$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{4x}^{\mathrm{2}} +\mathrm{6}\right)}\:+\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{4}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\right)}\:= \\ $$$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}−\mathrm{x}\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{6}}\:−\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }}\:= \\ $$$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}x}^{\mathrm{2}} \sqrt{\mathrm{4}+\frac{\mathrm{6}}{\mathrm{x}^{\mathrm{2}} }}−\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }}\:= \\ $$$$\:\mathrm{let}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{h}\:\wedge\:\mathrm{h}\rightarrow\mathrm{0} \\ $$$$\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}+\mathrm{6h}^{\mathrm{2}} }−\sqrt{\mathrm{4}+\mathrm{2h}^{\mathrm{2}} }}{\mathrm{h}^{\mathrm{2}} }\:=\:\mathrm{2}×\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{h}^{\mathrm{2}} }−\sqrt{\mathrm{1}+\frac{\mathrm{h}^{\mathrm{2}} }{\mathrm{2}}}}{\mathrm{h}} \\ $$$$=\:\mathrm{2}×\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{3h}^{\mathrm{2}} }{\mathrm{4}}\right)−\left(\mathrm{1}+\frac{\mathrm{h}^{\mathrm{2}} }{\mathrm{4}}\right)}{\mathrm{h}^{\mathrm{2}} }=\:\mathrm{1}\: \\ $$