Trying to prove the real-valued function has a limit. I am trying to show that the function $f:\mathbb{R}^2\to\mathbb{R}$ by the formula
$$f(\mathbf{x}) = \dfrac{x_1x_2^2}{x_1^4+x_2^2}  \textrm{ if }\mathbf{x} \neq \mathbf{0}$$
$$f(\mathbf{0}) = 0$$
has a limit of $0$ as $\mathbf{x}\to\mathbf{0}$. 
My book says I have to start by letting $\epsilon>0$ and find $\delta>0$ such that $||\mathbf{x}||<\delta$ implies $|f(\mathbf{x})|<\epsilon$. 
So I let $\epsilon>0$. $$|f(\mathbf{x})| < \epsilon$$
$$|\dfrac{x_1x_2^2}{x_1^4+x_2^2}|<\epsilon$$
I know I have to get $\delta$ in terms of $\epsilon$ so $\delta$ will be chosen sufficiently corresponding to $\epsilon$. I think I need to somehow use $||\mathbf{x}||<\delta \Longleftrightarrow \sqrt{x_1^2+x_2^2} < \delta$ and relate this equation to $|\dfrac{x_1x_2^2}{x_1^4+x_2^2}|<\epsilon$. I am lost and need help on how I would approach this problem next.
 A: If the value you get is independent from the x1 and y1 coordinates then it exists, no need to use the epsilon delta definition:
I am gonna assume x1=x and x2=y
Since x and y are in the (x,y) plane, I can write x and y using polar coordinates:
$$ x=rcos(\theta) $$
$$ y=rsin(\theta) $$
the limit will become:
$$ \lim_{r \rightarrow 0}\frac{r^{2}sin^{2}(\theta)rcos(\theta)}{r^{4}cos^{4}(\theta)+r^{2}sin^{2}(\theta)} $$
Rearranging:
$$ \lim_{r \rightarrow 0}\frac{r^{3}sin^{2}(\theta)cos(\theta)}{r^{2}(r^{2}cos^{4}(\theta)+sin^{2}(\theta))}$$
$$ \lim_{r \rightarrow 0}\frac{rsin^{2}(\theta)cos(\theta)}{r^{2}cos^{4}(\theta)+sin^{2}(\theta))}= \frac{0\times sin^{2}(\theta)cos(\theta)}{0 \times cos^{4}(\theta)+sin^{2}(\theta))}=0 $$
The limit exists as it is always zero, independent of x and y coordinates, if the value was dependent of theta the limit would not exist.
A: Mario's answer is good, but it is also good to practice $\varepsilon-\delta$ proofs. Let $x:=x_1$ and $y:=x_2$ for easy typing. Fix an arbitrary $\varepsilon>0$. First notice that for any $(x,y)\neq (0,0)$
\begin{align*}
|f(x,y)-0|& = \left| \frac{xy^2}{x^4+y^2}\right | \\
& \leq \frac{|x|y^2}{y^2}~\text{noting that}~ x^4,y^2\geq0~\text{,and assuming that}~y\neq 0 \\
&=|x| = \sqrt{x^2} \\
&\leq \sqrt{x^2+y^2} =\|(x,y)-(0,0)\|.
\end{align*}
If $y=0$ then $|f(x,y)-0|=0<|x|$, so our inequality holds for all $(x,y)\neq (0,0)$. Now choose $\delta=\varepsilon>0$. Then for any $(x,y)\in \mathbb{R^2}$ such that $0<\|(x,y)-(0,0)\|<\delta$ it follows from above that $|f(x,y)-(0,0)|<\varepsilon$. By the definition of a limit the result follows. We have found for any $\varepsilon>0$ a $\delta>0$, such that, for any point within a $\delta$ radius of the point, the difference between the value of the function at the point and the suspected limit is less than $\varepsilon$. 
I think it is important to note that having $f(\bar 0)=0$ is only important if you want to prove that $f$ is continuous at $\bar 0$. The whole point of a limit is that we don't care what happens at the point, only arbitrarily close to the point. 
A: For continuity at $(0,0)$:
If $x_2=0$ and $x_1\ne 0$ then $f(x_1,x_2)=0.$
If $x_2\ne 0$ then $|f(x_1,x_2)|\leq |x_1x^2_2|/|x_2^2|=|x_1|<\sqrt {x_1^2+x_2^2}\;=\|(x_1,x_2)\|.$
