Multivariable limit $\lim_{(x,y,z) \to (0,0,0)} \frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}$ I am studying multivariable limits from the text Vector Calculus by S.J. Colley.
I practiced computing limiting values of the functions of two-variables $f(x,y)$. 
Here's a problem involving a function of three variables. 

Compute the limit $$\lim_{(x,y,z) \to (0,0,0)} \frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}$$

I thought to myself, this could be for example, the scalar potential of a field in space. So, it's fun to see if the function has a limit as we get closer and closer to the origin. I would like to ask, if my proof is mathematically correct.
Solution.
From solid analytical geometry, a straight line in $\mathbb{R^3}$ is:
$\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}=r$
The straight line passing through the origin and having direction numbers $l,m,n$ is:
$\frac{x}{l}=\frac{y}{m}=\frac{z}{n}=r$
So, along this straight line with direction numbers $l,m,n$, our function takes on values
$f(x,y,z)=f(r)=\frac{2r^2l^2+3r^2m^2+r^2n^2}{r^2l^2+r^2m^2+r^2n^2}=\frac{2l^2+3m^2+n^2}{l^2+m^2+n^2}$
Along the line $x=y,z=0$, $f(x,y,z)=\frac{5}{2}$
Along the line $y=z,x=0$, $f(x,y,z)=\frac{4}{2}=2$
Along the line $x=z,y=0$, $f(x,y,z)=\frac{3}{2}$
Along the line $x=y=z$, $f(x,y,z)=\frac{6}{3}=2$
Hence, 
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }x=y,z=0}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=\frac{5}{2}$
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }y=z,x=0}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=2$
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }x=z,y=0}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=\frac{3}{2}$
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }x=y=z}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=3$
Thus, the limit of this function does not exist as $(x,y,z) \to (0,0,0)$.

Your limit was $\lim_{r\to0}\frac{2r^2l^2+3r^2m^2+r^2n^2}{r^2l^2+r^2m^2+r^2n^2}=\frac{2l^2+3m^2+n^2}{l^2+m^2+n^2}$ which depends on $(l,m,n)$ i.e. the direction of approach and hence the limit is path dependent.
You can see this by setting arbitrary values for $(l,m,n)\ne(0,0,0)$. For example, for $l=m=n=1$, your line of approach is $x=y=z$ and the limit is $6/3=2$. Along another line, say $\frac x2=\frac y2=z$, you get $21/9$. Hence the limit doesn't exist.
