- #1
Cristopher
- 9
- 0
I want to evaluate [itex]\displaystyle\lim_{(x,y)\to(-1,0)}\frac{y^4(x+1)}{|x+1|^3+2|y|^3}[/itex]
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.
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.