Question Number 216698 by sniper237 last updated on 16/Feb/25
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}\:−^{\mathrm{3}} \sqrt{{x}}\:\overset{?} {=}\:\mathrm{0} \\ $$
Answered by MrGaster last updated on 16/Feb/25
$$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\:^{\mathrm{3}} \sqrt{{x}}=\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\frac{\mathrm{1}}{\mathrm{3}}{x}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} +{O}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{2}} \right) \\ $$$$\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −{x}^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} +{O}\left({x}^{−\frac{\mathrm{5}}{\mathrm{3}}} \right)−{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} +{O}\left({x}^{−\frac{\mathrm{5}}{\mathrm{3}}} \right) \\ $$$$\underset{{x}\rightarrow\infty^{+} } {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} +{O}\left({x}^{−\frac{\mathrm{5}}{\mathrm{3}}} \right)\right)=\mathrm{0} \\ $$
Answered by mehdee7396 last updated on 17/Feb/25
$$\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{{x}}\right)=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{1}\right){x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow{li}\underset{{x}\rightarrow\infty} {{m}}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{{x}}\right)=\mathrm{0} \\ $$