Multivariable limit $\lim\limits_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2}$ The problem is:
$$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2}.$$
The tutor guessed it didn't exist, and he was correct. However, I'd like to understand why it doesn't exist.
I think I have to turn it into spherical coordinates and then see if the end result depends on an angle, like I've done for two variables with polar coordinates. I don't know how though.
I know $\rho = \sqrt{x^2+y^2+z^2}$ and $\theta = \arctan \left(\frac{y}{x} \right)$ and $\phi = \arccos \left( \frac{z}{\rho} \right)$, but how on earth do I break this thing up?
 A: Assume $x=y=z=t$ and $t\to0$, then the ratio converges to $\frac37$.
Assume $x=y=t$, $z=-t$ and $t\to0$, then the ratio converges to $-\frac27$. In both cases, $(x,y,z)\to(0,0,0)$ when $t\to0$. The two limits are not equal hence the limit when $(x,y,z)\to(0,0,0)$ does not exist.
A: Let's take the limit $z\to0$ first, getting
$$\lim_{x,y\to 0} \frac{xy}{x^2+4y^2}$$
Now consider what happens if you take the limit along $y=x$:
$$\lim_{x,y\to 0} \frac{xy}{x^2+4y^2} = \lim_{x\to0} \frac{x^2}{5x^2} = \frac{1}{5}$$
and along $y=2x$:
$$\lim_{x,y\to 0} \frac{xy}{x^2+4y^2} = \lim_{x\to 0} \frac{2x^2}{17x^2} = \frac{2}{17}$$
That's enough to tell you that the limit depends on direction.
A: If you want to rewrite to spherical coordinates, what you need is expressions for the rectangular coordinates in terms of the spherical ones, such as $x=\rho\cos(\theta)\sin(\phi)$, rather than the other way around. (By the way, the name of the letter $\rho$ is spelled "rho").
However, I don't think going to spherical coordinates is the most instructive approach here. It will be a lot of work that just obscures the real trick of this limit. Instead, just fix some nonzero point $P_0=(x_0, y_0, z_0)$ and imagine $(x,y,z)$ going towards $(0,0,0)$ along the ray through $P_0$:
$$(x,y,z)=(rx_0, ry_0, rz_0)$$
Insert that in your expression and simplify to see how it varies with $r$ for fixed $P_0$. That should tell you something rather interesting about the limit.
A: The limit for problems like these do not exist since the limit depends on the direction you approach. For the problem you have mentioned, say you approach $(0,0,0)$ along the direction $y = m_y x$ and $z = m_z x$, where $m_y$, $m_z$ are some constants, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \lim_{x \rightarrow 0} \frac{m_y x^2+2m_y m_z x^2 + 3m_z x^2}{x^2+4m_y^2 x^2+9 m_z^2 x^2}$$
Hence,
$$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \lim_{x \rightarrow 0} \frac{m_y + 2m_y m_z + 3m_z}{1 + 4 m_y^2 + 9 m_z^2} = \frac{m_y + 2m_y m_z + 3m_z}{1 + 4 m_y^2 + 9 m_z^2}$$
You can key in different values for $m_y$ and $m_z$ and you will get different values.
Below are some instances of the different limits you get by approaching in different directions.
If you tend to zero along the direction $(1,0,0)$ i.e. $m_y = m_z = 0$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = 0.$$
If you tend to zero along the direction $(1,1,0)$ i.e. $m_y = 1$ and $m_z = 0$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac15.$$
If you tend to zero along the direction $(1,0,1)$ i.e. $m_y = 0$ and $m_z = 1$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac3{10}.$$
If you tend to zero along the direction $(1,1,1)$ i.e. $m_y = 1$ and $m_z = 1$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac37.$$
