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$\lim_{(x,y)\rightarrow 0}\frac{xy^2}{x^2+y^4}$

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 enter image description here enter image description here enter image description here

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

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  • $\begingroup$ The graph region chosen seems maybe too large, and perhaps using a smaller region around the origin might show the limit 0.5. Could you re-sketch it and also include markings on the z axis? $\endgroup$ – coffeemath Feb 29 '16 at 1:42
  • $\begingroup$ @coffeemath I'm not very familiar with Mathematrica aout scaling $\endgroup$ – whoisit Feb 29 '16 at 2:07
  • $\begingroup$ whoisit-- try typing "? plot[options]" at maple command prompt. Generally options are for a given maple command. Maybe try "? Plot3D[options]" or just whatever you used to get your graph originally. The ? asks maple for its related definitions for commands, or their options. $\endgroup$ – coffeemath Feb 29 '16 at 2:13
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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 $(0,0)$.

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Since the math is impeccable I would suspect the fault lies with the values of $\Delta x$ and $\Delta y$ used by Mathematica. It has been years since I used Mathematica so I cannot advise you here. But if possible, set $\Delta x$ and $\Delta y$ to smaller values and see if you obtain a similar result.

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