What is a coordinate shifting? I need to find the limit:
$$\lim_{(x,y,z)\to(1,3-1)}\frac{(x-1)(y-3)+(z+1)^2}{(x-1)^2+2(y-3)^2+3(z+1)^2}.$$
A hint written below says: Perform a coordinate shifting to (0,0,0).
What does coordinate shifting means, and how should I use it in this case?
 A: The suggestion is to shift your coordinate system so that its origin is at the point $\langle 1,3,-1\rangle$. You can do this by subtracting $\langle 1,3,-1\rangle$ from each point, i.e., by applying the map
$$\langle x,y,z\rangle\mapsto\langle x-1,y-3,z+1\rangle\;.$$
You’ve done something similar when you observe that $y=(x-1)^2$ is just the parabola $y=x^2$ shifted $1$ unit to the right: replacing $y=(x-1)^2$ by $y=x^2$ is moving the origin of your coordinate system to $\langle 1,0\rangle$.
Another way to think of it: as $\langle x,y,z\rangle\to\langle 1,3,-1\rangle$, clearly $x\to 1$, $y\to 3$, and $z\to -1$, so $x-1\to 0$, $y-3\to 0$, and $z+1\to 0$. If you let $u=x-1$, $v=y-3$, and $w=z+1$, then $\langle u,v,w\rangle\to\langle 0,0,0\rangle$ as $\langle x,y,z\rangle\to\langle 1,3,-1\rangle$, and you’re looking at
$$\lim_{\langle u,v,w\rangle\to\langle 0,0,0\rangle}\frac{uv+w^2}{u^2+2v^2+3w^2}\;.$$
A: Coordinate shifting:
$$\text{By putting }\;\; u=x-1,\,\;v=y-3,\; \text{ and } \;w=z+1,$$ 
$$\text{then as }\,(x,y,z)\to (1,3,-1),\, \text{ we have }\,(u,v,w)\to(0,0,0)$$
This is essentially what coordinate shifting then gives us: it changes your limit (left-hand side) into the following equivalent limit (right-hand side):
$$\lim_{(x,y,z)\to(1,3,-1)}\;\frac{(x-1)(y-3)+(z+1)^2}{(x-1)^2+2(y-3)^2+3(z+1)^2}\; = \;\lim_{(u, v, w)\to(0,0,0)}\;\frac{uv+w^2}{u^2+2v^2+3w^2}.$$
A: It means you create new positions(coordinates) so that the expression is simplified.
Put x-1=X, y-3=Y etc. ,so that when x-->1, X-->0. Thus you shift the coordinates. In this case x is shifted back 1 unit. 
