3-Variable Limit $\lim\limits_{(x,y,z,) \to (0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$ $$\lim_{(x,y,z,) \to (0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$$
Background: I think that converting the formula into parametric variables won't work since I only know that's useful for converting $\sin^2(x) + \cos^2(x)$ into $1$. I've actually never done a limit with three variables before so I'm a little confused about how we can go about it. Any suggestions as to how I should begin to approach this?
 A: Approach along $x=y=z=a$:
$$\lim_{(x,y,z)\to(0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}=\lim_{a\to0}\frac{a^2+2a^3}{2a^2+a^4}=\lim_{a\to0}\frac{1+2a}{2+a^2}=\frac12$$
Approch along $x=z=a$, $y=a^2$:
$$\lim_{(x,y,z)\to(0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}=\lim_{a\to0}\frac{2a^3+a^4}{a^2+2a^4}=\lim_{a\to0}\frac{2a+a^2}{1+2a^2}=0$$
Hence, the limit doesn't exist.
A: Plugging $x=\frac{1}{n},y=\frac{1}{n^2},z=\frac{1}{\sqrt{n}}$ in your term you got the limit $\frac{1}{2}$.
Using $x=\frac{1}{n^3},y=\frac{1}{n^4},z=\frac{5}{\sqrt{n}}$ you will get $0$. Thus no limit exist.
A: limit of the equation does not exist . take two cases 
$$1)y=z=x $$   then limit is $$(x^2+2x^3)/2x^2+x^4$$  where x tends to zero so you get 1/2 now similarly take $$2)y=2x,z=x$$ then limit is $$(2x^2+3x^3)/5x^2+x^4$$ so you get 2/5 which is different from 1/2 . so no limit exists
A: Using the substitution $z^2 = t$ we get 
$$\lim_{(x,y,z,) \to (0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}= \lim_{(x,y,t,) \to (0,0,0_{+})} \frac{xy+yt+xt}{x^2+y^2+t^2}$$
The expression $\frac{xy+yt+xt}{x^2+y^2+t^2}$ is homogenous of degree $0$ and therefore takes a constant value on each line through the origin. Since these values are not constant as the line varies there is no limit. 
