Problems with the existence of limits of several variables I started some time studying calculus of two variables, and I'm having difficulty in knowing (and prove) that a limit does not exist, how could I resolve, for example, these, whose statement asks to show that there are? $$\lim_{(x,y)\to(0,0)} \frac{x^9y}{(x^6+y^2)^2}\\\lim_{(x,y)\to(0,0)} \frac{x^2}{x^2+y^2}\\\lim_{(x,y,z)\to(0,0,0)} \frac{x^3+yz^2}{x^4+y^2+z^2}\\\lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^4+y^4}\\\lim_{(x,y,z)\to(0,0,0)} \frac{x^2+y^2-z^2}{x^2+y^2+z^2}$$
 A: You need consider paths on the $xy$-plane as follows:
If a limit exists, then it must approach the same value no matter what way you approach $(0,0)$. Consider the first example. If you take a line $y=x$ and approach $(0,0)$ along that line, you get,
$$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^9y}{(x^6+y^2)^2} =  \lim_{x\rightarrow 0} \frac{x^{10}}{(x^6 + x^2)^2}$$
Does this limit exist? Now try with $y = x^3$.
Similarly for the second problem, consider the path $y = mx$, and you get
$$\lim_{(x,y)\rightarrow (0,0)} \frac{x^2}{(x^2+y^2)} = \lim_{x\rightarrow 0} \frac{x^2}{x^2 + m^2x^2} = \frac{1}{1+m^2}$$
So the limit depends on $m$. Hence the limit cannot exist. Can you think of similar paths for the remaining examples?
A: 1) What happens if we approach along $y=x$? What about if we approach along $y=x^3$?
2) For this one, there is a "universal" strategy: if we see $x^2+y^2$ at the bottom, then $x=r\cos\theta$, $y=r\sin\theta$ is almost always useful Simplify, we get $\cos^2\theta$. Thus what happens depends on the direction of approach, the limit does not exist.
3) If we set $y=z=0$, the thing blows up (in different directions) as $x\to 0$.  
4) We get different limits if we approach along $y=x$ and if we approach along $y=2x$.
5) Set $z=0$ and see what happens. Set $x=0$ and $y=z$ and see what happens. 
