Question Number 143013 by Study last updated on 08/Jun/21
$${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$
Commented by Study last updated on 08/Jun/21
$${what}\:{is}\:{the}\:{practice}? \\ $$
Commented by Dwaipayan Shikari last updated on 08/Jun/21
$$\mathrm{0} \\ $$
Commented by Dwaipayan Shikari last updated on 08/Jun/21
$$\frac{\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}{{x}+\mathrm{1}}\:\:=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} }\:\: \\ $$$${x}\:{is}\:{very}\:{large}\:{so}\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} }\rightarrow\mathrm{0} \\ $$