Question Number 54992 by Mikael_Marshall last updated on 15/Feb/19
$$\underset{{n}\rightarrow\infty} {{lim}}\left(\sqrt{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}−{n}\right) \\ $$
Answered by kaivan.ahmadi last updated on 15/Feb/19
$$×\frac{\sqrt{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}+{n}}{\sqrt{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}+{n}}={li}\underset{{n}\rightarrow\infty} {{m}}\frac{{n}^{\mathrm{3}} +{n}+\mathrm{1}}{\sqrt{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}+{n}}\approx \\ $$$${li}\underset{{n}\rightarrow\infty} {{m}}\frac{{n}^{\mathrm{3}} }{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }=\infty \\ $$$$ \\ $$
Answered by tm888 last updated on 16/Feb/19
$$=\infty \\ $$