# Limit multivarible

How do I solve this limit?

$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz}$$

This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$

The first one is a standard limit with value one, but I'n not sure about the other term.

• change to spherical coordinates. It'll become $\lim_{\rho\to 0}\frac{\sin(\rho^2)}{\rho^2+\rho^3 A(\theta)B(\phi)}$ – PNDas Oct 30 at 11:15
• What is the next step? – Erika Oct 30 at 11:17
• I made some changes, can you solve it now? – PNDas Oct 30 at 11:20

Yes your idea is very good to put away the $$\sin$$ term, now we have that, for example by spherical coordinates

$$\frac{xyz}{x^2+y^2+z^2} =r \cos \theta\sin \theta\cos^2 \phi\sin \phi$$

or as an alternative by AM-GM we can use that

$$x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2y^2z^2} \implies \left|\frac{xyz}{x^2+y^2+z^2}\right| \le \frac{|xyz|^\frac13}{3}$$

concluding in both cases by squeeze theorem.

• Yeah, got it. Thanks – Shubham Johri Oct 30 at 11:36

$$0 \leq | \frac {xyz} {x^{2}+y^{2}+z^{2}}|\leq \frac 1 2 |z| \to 0$$ since $$|xy| \leq \frac 1 2{(x^{2}+y^{2})}$$. Hence the given limit is $$1$$.