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Does this $$\lim_{x,y,z\to(0,0,0)}\frac{xy+2xz+yz}{{x^2+y^2+z^2}}$$ have a limit?

My answer for this is Let f(x,y,z)=$$\frac{xy+2xz+yz}{{x^2+y^2+z^2}}$$ then, $$\lim_{x\to0}{f(x,0,0)}=\lim_{x\to0}\frac{0}{x^2}=0$$

$$\lim_{x\to0}{f(x,x,0)}=\lim_{x\to0}\frac{x^2}{2x^2}=\frac{1}{2}$$

Since this two limit are not the same,$$\lim_{x,y,z\to(0,0,0)}\frac{xy+2xz+yz}{{x^2+y^2+z^2}}$$ does not exist.

I'm not sure if this justification is enough or correct.

Karen
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    Yes, this is fine, except that you should have $\frac0{x^2}$ in the first limit. – Brian M. Scott Jun 11 '13 at 07:51
  • Oh thanks! I missed that. So, this is fine to do in exams? Because my lecturer showed a different way but I am totally at loss at it except for this... – Karen Jun 11 '13 at 07:54
  • @Karen : Sorry I deleted my comment, I was about to copy paste in an answer but I pressed CTRL+V instead of CTRL+C... There are probably many ways to show this limit doesn't exist indeed ; assuming it exists can lead to many different contradictions, not just one. For instance, you could also consider the path $(x,x,x)$ and the limit would be $4/3$. :) – Patrick Da Silva Jun 11 '13 at 07:55
  • It certainly ought to be fine: you’ve shown that the limit when you approach the origin along the $x$-axis is different from the limit when you approach it along the line $y=x$ in the $xy$-plane, and that certainly shows that the limit doesn’t exist. (And while you certainly don’t need any more, I’ll note that if you approach along the line $x=y=z$, the limit is yet a third number, $\frac43$.) – Brian M. Scott Jun 11 '13 at 07:55
  • oh i see! Thank you guys! – Karen Jun 11 '13 at 08:00
  • You’re welcome! – Brian M. Scott Jun 11 '13 at 08:01

3 Answers3

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In light of @Brian's comment you can take the following path as well:

$$r_{\alpha,\beta}(t)=(t,\alpha t,\beta t^2),~~~ \alpha,\beta\in \mathbb R$$

Mikasa
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Recalling spherical coordinates

$$ x = \rho \cos(\theta)\sin(\phi),\, y=\rho \sin(\theta)\sin(\phi),\, z =\rho \cos(\phi), \quad 0\leq \theta \leq 2\pi,\, 0\leq \phi \leq \pi. $$

we have

$$ \frac{xy+2xz+yz}{{x^2+y^2+z^2}}= \cos(\theta)\sin(\theta)\sin^2(\phi)+2\cos(\theta)\sin(\phi)\cos(\phi)+ \sin(\theta)\sin(\phi)\cos(\phi). $$

Now the above expression achieves an infinite number of values depending on $\theta$ and $\phi$ which implies the limit does not exist.

Mhenni Benghorbal
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Similar to your solution, with some geometry.
Let us consider the $(x,y)$-plane; then $f(x,y,0)=\frac{xy}{x^2+y^2}$. The limit

$\lim_{(x,y)\rightarrow (0,0)} f(x,y,0)$

can be solved using polar coordinates, i.e. $x=r\cos\theta$ and $y=r\sin\theta$ with $r>0$ and $\theta\in[0,2\pi)$. But then

$\lim_{(x,y)\rightarrow (0,0)} f(x,y,0)=\lim_{r\rightarrow 0} f(r\cos\theta, r\sin\theta)$,

with $f(r\cos\theta, r\sin\theta)=\cos(\theta)\sin(\theta)$. The above limit does not exist as it depends on the direction (represented by a choice of $\theta$) we choose to arrive at $(0,0,0)$.

Avitus
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