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.

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 $|f(x_1,x_2)|\leq |x_1x^2_2|/|x_2^2|=|x_1|<\sqrt {x_1^2+x_2^2}\;=\|(x_1,x_2)\|.$
