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I would like to find the following limit $$\lim_{(x,y,z)\to(0,0,0)}\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}.$$

It looks like it would be zero since if we put $M=\max\{x,y,z\}$ and $m=\min\{x,y,z\}$, then $$\Big|\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}\Big| \leq \Big|\frac{M^5}{m^4}\Big|$$ so the exponent in the denominator is bigger than the exponent in the numerator.

But how do I actually calculate this limit?

Thank you.

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1 Answer 1

up vote 4 down vote accepted

Let $r = \sqrt{x^2+y^2+z^2}$. Then, using the generalized mean inequality, we have:

$\sqrt[4]{\dfrac{x^4+y^4+z^4}{3}} \ge \sqrt[2]{\dfrac{x^2+y^2+z^2}{3}} \ge \sqrt[3]{|xyz|}$.

Therefore, $x^4+y^4+z^4 \ge 3\left(\dfrac{x^2+y^2+z^2}{3}\right)^2 = \dfrac{r^4}{3}$ and $|xyz| \le \left(\dfrac{x^2+y^2+z^2}{3}\right)^{3/2} = \dfrac{r^3}{3\sqrt{3}}$.

Hence, $\left|\dfrac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}\right| = \dfrac{|xyz| \cdot |x^2+y^2+z^2|}{|x^4+y^4+z^4|} \le \dfrac{\frac{r^3}{3\sqrt{3}} \cdot r^2}{\frac{r^4}{3}} = \dfrac{r}{9\sqrt{3}} = \dfrac{\sqrt{x^2+y^2+z^2}}{9\sqrt{3}}$.

Now, apply the squeeze theorem.

If you don't like this approach, you can try using spherical coordinates. Let $x = \rho \sin\phi \cos\theta$, $y = \rho\sin\phi\sin\theta$, and $z = \rho\cos\phi$. After making those substitutions, the function will simplify to $\rho \cdot \left(\text{some trig expression with} \ \phi \ \text{and} \ \theta \ \text{that I hope is bounded}\right)$, which will approach $0$ as $\rho \to 0$.

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Fantastic answer, thank you! Could you please demonstrate the approach with spherical coordinates? I'm not sure how to do it, but I'd love to see that as well. –  rehband Aug 18 at 8:03
    
The problem is that it never says $x,y,z$ converge to zero at the same order. –  Troy Woo Aug 18 at 8:03
1  
^Yes, that is precisely what makes multivariable limits difficult to evaluate. In this case, if $(x,y,z)$ is within a distance of $r$ of the origin, then the magnitude of the function is at most $\frac{r}{9\sqrt{3}}$. Formally, we can let $\epsilon > 0$ be arbitrary and choose $\delta = 9\sqrt{3}\epsilon$. If $\|(x,y,z)-(0,0,0)\| = \sqrt{x^2+y^2+z^2} < \delta$, then by the above math $|f(x,y,z) - 0| < \epsilon$. –  JimmyK4542 Aug 18 at 8:12

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