Use the definition of the limit to verify $\lim\limits_{(x,y,z) \to 0} \frac {(5x^3y^2+6xz^4)}{(2x^2+3y^2+z^2)}$

Question

$$\lim\limits_{(x,y,z) \to 0} \frac {(5x^3y^2+6xz^4)}{(2x^2+3y^2+z^2)}=0 $$

I'm having a problem how to start this since it's new for me. I know that I have to isolate R but not sure how.

Answer 1

It is easy to verify that $|5x^{3} y^{2} +6x z^{4}| \leq |x| |(2x^{2}+3y^{2}+z^{2})| |3y^{2}+6z^{2}|$ and the result follows from this. Given $\epsilon >0$ take $0<\delta <1/9$. You can now verify that the given expression is less than $\epsilon$ in absolute value when $\sqrt (x^{2}+y^{2}+z^{2}) <\delta$

Answer 2

Using spherical coordinates $$ x = r\sin\theta\cos\phi, \ y = r\sin\theta\sin\phi, \ z = r\cos\theta $$

we get $$ \frac{5x^3y^2 + 6xz^4}{2x^2 + 3y^2 + z^2} = r^3\cdot \frac{5\sin^5\theta\cos^3\phi\sin^2\phi + 6\sin\theta\cos^4\theta\cos\phi}{2\sin^2\theta\cos^2\phi + 3\sin^2\theta\sin^2\phi + \cos^2\theta} $$

In the limit of $r\to 0$, the above expression always goes to $0$, independent of $\theta, \phi$

Answer 3

Note that

$$0\le \frac {|5x^3y^2+6xz^4|}{2x^2+3y^2+z^2}\le \frac{|5x^3y^2+6xz^4|}{x^2+y^2+z^2}= \rho^3\cdot (5\sin^5\theta\cos^3\phi\sin^2\phi + 6\sin\theta\cos^4\theta\cos\phi)\to0$$

thus by squeeze theorem

$$\frac {|5x^3y^2+6xz^4|}{2x^2+3y^2+z^2}\to 0\iff \frac {5x^3y^2+6xz^4}{2x^2+3y^2+z^2}$$