- #1

Cristopher

- 9

- 0

With some help, I was able to prove that the limit is 0, using Hölder's inequality. Like this:

[tex]\left(|x+1|^3\right)^{1/5}\left(\frac{1}{2}|y|^3\right)^{4/5}\leq\frac{1}{5}|x+1|^3+\frac{4}{5}\frac{1}{2}|y|^3[/tex]

Raising to the [itex]5/3[/itex] power and cancelling we get:

[tex]|x+1|\left(\frac{1}{2}\right)^{4/3}y^4\leq\left(\frac{1}{5}\right)^{5/3}\left(|x+1|^3+2|y|^3\right)^{5/3}\\

\frac{|x+1|y^4}{|x+1|^3+2|y|^3}\leq\sqrt[3]{\frac{16}{3125}}\left(|x+1|^3+2|y|^3\right)^{2/3}[/tex]

But I wonder if there are any other ways to prove it. Does anyone have other ideas?

Thanks for any input.