limit of $\frac{2xy^3}{7x^2+4y^6}$, different answers. Good evening, everyone I've tried my possible best to evaluate the limit as $(x,y) \to (0,0)$ but Using sage the answer is 0 either ways but one textbook is saying the limit doesn't exist.  Could the 2 answers be correct?
 A: This limit does not exist.
If you approach $(0,0)$ along the $x$-axis, your expression tends to $0$.
On the other hand, if you approach along the curve $x=y^3$, you get
$$
\lim_{\substack{(x,y)\rightarrow(0,0)\\\text{along $x=y^3$}}}\frac{2xy^3}{7x^2+4y^6}=\lim_{x\rightarrow0}\frac{2y^6}{11y^6}=\frac{2}{11}\neq0.
$$
A: The Wolfram|Alpha result is indeed a weakness (bug).
In slightly more detail, which I hope I won't have to regret, the underlying problem is in Mathematica's Limit handling of intervals.
ee = (rho*Interval[{-1, 1}])/(rho^2*Interval[{-1, 1}] + 
     Interval[{-1, 1}]);
In[63]:= Limit[ee, rho -> 0]
This incorrectly gives zero when it should return unevaluated.
This is a known bug. What's not known at this point is how to fix it.
A: Take $y=\sqrt[3]{x}$, then you obtain
$$\frac{2xy^3}{7x^2+4y^6}=\frac 2{11}\neq 0$$.
A: My guess is the problem is that sage cannot test every approach curve, so it probably tests a small grid of points about the origin. Since the place where the limit is bad is shaoed like a curved cubic, the grid will miss the curve and give the wrong answer.
Note that Wolfram Alpha has the same problem: http://m.wolframalpha.com/input/?i=limit+as+x+goes+to++0%2C+y+goes+to+0+of+%5Cfrac%7B2xy%5E3%7D%7B7x%5E2%2B4y%5E6%7D&incCompTime=true
