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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.

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

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  • $\begingroup$ Yeah, got it. Thanks $\endgroup$ – Shubham Johri Oct 30 at 11:36
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$0 \leq | \frac {xyz} {x^{2}+y^{2}+z^{2}}|\leq \frac 1 2 |z| \to 0$ since $|xy| \leq \frac 1 2{(x^{2}+y^{2})}$. Hence the given limit is $1$.

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