match limit of different paths on a graph $\lim_{(x,y)\rightarrow 0}\frac{xy^2}{x^2+y^4}$
if we consider the path $x=y^2$ and get
$\lim_{(y^2,y)\rightarrow 0}\frac{y^2(y^2)}{(y^2)^2+y^4}=\lim_{y\rightarrow 0}\frac{y^4}{2y^4}=\frac{1}{2}$
but if I used Mathematrica to draw a graph of this function



it looks like the graph doesn't match the limit should be 0.5 when follow the path $x=y^2$,

For
$
\lim_{(x,y)\rightarrow (0,0)}\frac{xy^2}{x^2+y^4},
$
if you set $x = -y^2$, the limit is
$$
\lim_{y\rightarrow 0}\frac{(-y^2)y^2}{(y^2)^2+y^4} = -\frac 12,
$$
and if you set either $x = 0$ or $y = 0$, the limit is
$$
\lim_{y\rightarrow 0}\frac{0(y^2)}{0^2+y^4}
= \lim_{x\rightarrow 0}\frac{x(0^2)}{x^2+0^4} = 0.
$$
Some 3D plotting software, given an arbitrary function over $x$ and $y$, computes the $z$-values at some finite number of $(x,y)$ values in
a grid pattern. If the granularity of the grid is $\delta$ units,
the software will compute the function at points along the $y$-axis,
and it will compute it at points along
the lines $x=\delta$ and $x=-\delta$, but it will not compute it at points
that are closer than $\delta$ to the $y$-axis while not being exactly
on the $y$-axis.
The points that form the "ridges" at heights $z=\frac12$ and $z=-\frac12$
are along the curves $x = y^2$ and $x = -y^2$, which get very close to the
$y$ axis while they are still relatively far from the $x$ axis.
If the plot is done according to a grid of $(x,y)$ values,
inevitably you will "lose" part of the ridges where they get too close
to the $y$-axis to show up as plotted values.
It seems a reasonable guess that Mathematica is doing something like that.
If you can get it to use a smaller grid, you can make the gap a bit smaller.
But in order to make the gap half as large, you must make the
grid size one quarter as large.
In general, I think you're likely to have problems like this with any
function that is bounded as $(x,y)\to(0,0)$ but that does not have
a (path-independent) limit at $(0,0)$.
