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Possible error? Simple Limit discrepancy between W|A and Mathematica

Agustin Covarrubias
Posted 6 years ago

The error

It's very straight-forward; the same piece of code (presumably interpreted the same way between both) outputs a different answer. I did the limit with polar coordinates and the answer is, indeed, 1/4. If you know what may have happened here, let me know.
POSTED BY: Agustin Covarrubias
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Have you guys at Wolfram recently created methods for multivariate limits? When?
POSTED BY: Agustin Covarrubias

Possibly Alpha is running an older version that does not have more recent methods for handling multivariate limits.
POSTED BY: Daniel Lichtblau

Thanks Bill, I have just happened to actually stumble upon your reply from 4 months ago to another person which had a somewhat similar issue, and typing "limit calculator" and adding the second variable from the menu seemed to help. Although it showed the correct answer (1/4) it gave me back only 2 directions from which W|A evaluated the limit, which means it's not necessarily correct, given the particular nature of multivariate limits.

I'm still concerned about Wolfram|Alpha not being able to do this right 100% though.
POSTED BY: Agustin Covarrubias
Bill Nelson
Bill Nelson
Posted 6 years ago

While it is true that Wolfram|Alpha can interpret some Mathematica syntax, it is not able to interpret all Mathematica syntax and that presumably means Mathematica syntax is not all being interpreted in exactly the same way by Mathematica and Wolfram|Alpha.

In examples I have seen and tried Wolfram|Alpha does seem to correctly evaluate limits involving a single variable. If you find any counterexample to that then please make that known.

I have seen examples in the past where multivariate limits do not seem to all be understood and supported by Wolfram|Alpha, as you seem to have demonstrated.
POSTED BY: Bill Nelson

I pasted the query once again on the W|A web app, and it still returns the same answer query

Thanks for your answer, but my main concern is not about the limit itself, which can be easily calculated via the polar coordinates method I mentioned. My concern is that the same code (presumably interpreted the same way) gave back very different answers between Wolfram Mathematica and W|A.
POSTED BY: Agustin Covarrubias
POSTED BY: Gianluca Gorni 
Limit[(x^2*(1 + Sin[y]) + y^2)/(4 x^2 + 4 y^2), {x, y} -> {0, 0}]
POSTED BY: Agustin Covarrubias

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POSTED BY: EDITORIAL BOARD
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