Find limits of a function with several variables 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. 
 A: In light of @Brian's comment you can take the following path as well:
$$r_{\alpha,\beta}(t)=(t,\alpha t,\beta t^2),~~~ \alpha,\beta\in \mathbb R$$
A: 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. 
A: 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 $(0,0,0)$.
