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How to calculate $$\lim _{( x,y) \rightarrow (0,0) }\dfrac {x^2y^{3} }{\sin ( x^{8}+y^{4}) }$$

Wolfram Alpha says it is equal to $0$, but if calculate by trying $y=x$ then I get the limit $\lim \limits_{ x \to 0 }\dfrac {x^5 }{\sin ( x^{8}+x^{4}) }$ which it says it's equal to $0$.

Now if I try $y=x^2$ then I get the limit $\lim\limits_{ x \to 0 }\dfrac {x^8 }{\sin ( 2x^{8}) }$ which is equal to $\frac12$.

Which one is true, is there no limit or does limit equal to $0$?

Winther
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memonto
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1 Answers1

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This is a good example for why one should not blindly trust Wolfram Alpha when it comes it multidimensional limits. Such limits are hard to do for a computer as the result can depend on the path taken towards the limit-point and there are infinitely many paths to approach any given point. When the limit does not exist the standard solution method is to find different paths that leads to different limits. What works in any given case if often based on experience, i.e. just 'seeing' what are some good paths to try. This type of approach is hard to implement as an algorithm on a computer.

Wolfram Alpha gets this one wrong. To see this write the limit as

$$\lim_{(x,y)\to (0,0)} \frac{x^2y^3}{x^8+y^4} \cdot \frac{x^8+y^4}{\sin(x^8+y^4)} = \lim_{(x,y)\to (0,0)} \frac{x^2y^3}{x^8+y^4} \cdot \lim_{(x,y)\to (0,0)} \frac{x^8+y^4}{\sin(x^8+y^4)}$$

The latter limit is $1$ as $x^8+y^4\to 0$ as $(x,y)\to 0$ and $\lim\limits_{z\to 0}\frac{z}{\sin(z)} = 1$. The former limit does not exist as we can see by taking the limit along curves $x=k\sqrt{y}$ then

$$\lim_{(x,y)\to (0,0)} \left.\frac{x^2y^3}{x^8+y^4}\right|_{x=k\sqrt{y}} = \lim_{y\to 0} \frac{k^2y^4}{k^8y^4 + y^4} = \frac{k^2}{k^8 + 1}$$

and since the limit is path dependent (it depends on the value we take for $k$) it follows that $\lim\limits_{(x,y)\to (0,0)} \frac{x^2y^3}{x^8+y^4}$ does not exist.

Winther
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  • Your raisonnement looks correct, so may be I'm wrong (because I get $0$ for limit). Could you check it ? May be you'll find a mistakes, and if you find a mistakes, I will erase my answer. – Surb Nov 12 '15 at 18:32
  • I dont understand how wolfram miscalculates this kinda limits. It can easily detect multidimentional limits and try numerous path in a little time and compare them if they are equal. it sounds doable to me (: – memonto Nov 12 '15 at 18:32
  • @memonto I can only speculate, but I suspect it just computes $\lim_{x\to 0}\lim_{y\to 0}$ and $\lim_{y\to 0}\lim_{x\to 0}$ and then concludes from this (you could try asking this at the Mathematica Stack Exchange). There are infinitely many paths to take and there is no obvious way to do this: for example $y = kx^n$, $y=e^{-1/x^n}$, $\ldots$ for all possible real numbers $n$. – Winther Nov 12 '15 at 18:42
  • Yes, thank you :-) – Surb Nov 12 '15 at 20:09
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    I've actually worked on this exact code and it tries to do quite a lot more than just both double limits. Nonetheless, you are absolutely correct that multivariate limits are particularly tricky in computer algebra in general and buggy in WolframAlpha in particular. Also, this question would certainly be quickly closed at mathematica.se as off topic. – Mark McClure Nov 12 '15 at 20:24
  • @MarkMcClure I was about to ask it there, but I noticed that WA questions was off-topic so thank you for leaving a comment here. It would be nice to know more about how WA does this so if you have a reference that explains this (or where algorithms for multidim limits in general is discussed) that would be great. – Winther Nov 12 '15 at 20:29
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    I don't have such a reference. If you're genuinely curious about how to go about it, a mathematica.se question that asks how one might go about it with Mathematica, without specific reference to WolframAlpha, would be quite reasonable. You might be some very good answers. DanL is certainly the expert on that particular topic. – Mark McClure Nov 12 '15 at 20:35