# Can wolframAlpha be wrong on this vector limit?

We had a homework on multivariable analysis, and there was this problem and the teacher said that we didnt trust wolfram but I'm not convinced on it, because of this.

Is $f(x,y)=\frac{x^2}{x^2+y^2-x}$, Continuous on (0,0), if not say what kind of discontinuity is it.

Clearly $f(0,0)=\frac{0^2}{0^2+0^2-0}$, its a form of indeterminate. So we go to the limit. $\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2+y^2-x}$ I get $0$, on some few cases, but i cant prove that its $0$, but i "asked" wolfram and he said its $0$, but some other of my class mates say that wolfram gave a non existing limit, or when they refreshed the site, it gave a different answer(which i think its very odd)

Is wolfram possibly wrong, or the limit there is $0$. I got a little far on proving that the limit does exist but, i could be wrong, because i cant finish it.

Any ideas on that limit?

• Consider the $k$-paths $t\mapsto (kt^2,t)$. Wolfram Alpha is awful for multivariable limits, but when it says the limit doesn't exist, it usually is the case. I at least have never seen it fail in this case. When it says it exists, though, tread carefully. – Git Gud Feb 27 '15 at 22:26
• what's the value of $f$ on $(0,0)$? – Ab urbe condita Feb 27 '15 at 22:28
• @GitGud: It fails in both ways. The programmer behind it just dumps in a whole bunch of heuristics which may either give a limit when it does not exist or claim non-existence when it does. See i.stack.imgur.com/bxUZ2.png for JUST ONE example. After they saw me give this example just last month, they changed the program and now it gives two contradictory answers. Any proper mathematician can with some creativity produce a never-ending list of counter-examples that Wolfram Alpha will get wrong! – user21820 Apr 5 '17 at 14:35
• @user21820 "Any proper mathematician". Ouch! Thanks, though $\ddot \smile$ – Git Gud Apr 15 '17 at 23:17
• @GitGud: Sorry if it came off the wrong way! I simply meant that it's just a matter of creativity to trick WA, since WA is not doing proper mathematics. =) – user21820 Apr 16 '17 at 16:37

When $x = y^2$, $f(x,y) = 1$, while when $x = 0$, $f(x,y) = 0$. Hence the limit doesn't exist as $(x,y) \rightarrow (0,0)$.