Continuity of this function as $(x,y)$ tends to $(0,0)$ 
Here's a function in $x$ and $y$ defined piecewise as
$$f(x,y)=
\left\{\begin{array}{ll}
      0 & (x,y)=(0,0)\\
 \frac{x^2y}{x^4+y^2} & (x,y) \neq (0,0) \\
\end{array}\right.$$
Examine its continuity as the ordered pair tends to $(0,0)$.

Okay, so I first tried this by assuming $x=\frac{1}{n}$ and $y=\frac{1}{n^2}$, where $n\rightarrow \infty$ . The limit of the function as $(x,y) \rightarrow (0,0)$ came out to be $\frac{1}{2}$, and since this is not equal to the value of the function at the said point, its discontinuous at the origin.
But when I assumed $x=\frac{1}{n}$  and $y=\frac{1}{n}$ , I got the limit zero:$$f(\frac{1}{n},\frac{1}{n})= \frac{\frac{1}{n^3}}{\frac{1}{n^4}+ \frac{1}{n^2}} = \frac{\frac{1}{n}}{\frac{1}{n^2}+ 1} $$ Since $x,y \rightarrow 0$,$f(\frac{1}{n},\frac{1}{n}) \rightarrow \frac{0}{0+1}=0$
Where am I going wrong?

Notice that :
$$f(t, t^2) = \dfrac{t^2 t^2}{t^4 + t^2} = \dfrac{t^4}{2 t^4} = \dfrac{1}{2} \underset{t \to 0}{\to} \dfrac{1}{2} \neq f(0, 0)$$
