Wolfram Alpha gives me: $\lim\limits_{(x,y)\to(0,0)} \dfrac{{xy^3}}{{x^2+y^6}}=0$.
I assume $f\left( {x,y} \right) = \dfrac{{x{y^3}}}{{{x^2} + {y^6}}}$
I find that when $x=y^3$, $f\left( {x,y} \right) = \dfrac{1}{2}$, while when $x=0, f\left( {x,y} \right) = 0$. Hence the limit does not exist.
Please correct me if I am wrong. Thank you.
Best Answer
Mathematica, and consequently WolframAlpha, does not have built-in capability to evaluate arbitrary multivariate limits. Therefore, the command
gives $0$, but
yields $1/2$. The path-dependence of the limit can only be handled when a path is specified. This is something that many users of WolframAlpha misunderstand about
Limit. If you writeMathematica will simply thread
Limit[]over the list of replacement rules{x -> 0, y -> 0}, producing the output $\{0,0\}$. To get a better understanding of what it's doing, try the commandClearly, it doesn't behave the way one might think it should behave. And if you think you can write
you get an error, because
{x,y}is not a valid variable forLimit. This error definitely shows that such multivariate limits are not within the scope of Mathematica's functionality.Now, that doesn't mean that WolframAlpha can't be a bit smarter than Mathematica's built in
Limit: if you enterhttp://www.wolframalpha.com/input/?i=Limit+of+x+y%2F(x%5E2%2By%5E2)+as+(x,y)+tends+to+(0,0)
You will get the proper response, but if you enter
http://www.wolframalpha.com/input/?i=Limit+of+x+y%5E3%2F(x%5E2%2By%5E6)+as+(x,y)+tends+to+(0,0)
you get a wrong response, because again, WolframAlpha is built on Mathematica's functionality. While it can handle some special queries, I suspect that these are probably pre-programmed somehow, or that they were special (but common) queries that were added to the knowledge base. In any case, I would not rely on WolframAlpha or Mathematica to handle multivariate limits.