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Why does Maple 14 gives me back the whole function when I want to use "limit" with two unknown variables?

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why the negatives? – gasko peter Mar 17 '13 at 18:14
Why the close votes? – Thomas Mar 17 '13 at 19:11
up vote 2 down vote accepted

Giving back the function normally indicates it can't compute the limit (and doesn't know if it exists or not), maple 16 gives the same result (none) which i find suprising as the limit is obvious $1$.

When $x\to 0$ and $y\to 0$ we know that $x^2+y^2\to 0$ so it is the same as $w\to 0$ with $w=x^2+y^2$ so we have the limit $$\lim_{w\to 0}\frac{\sin(w)}{w}=1$$

We know that $$\sin(w)=w-\frac{w^3}{3!}+\frac{w^5}{5!}\mp \dots$$ So $$\lim_{w\to 0}\frac{\sin(w)}{w}=\lim_{w\to 0}\left(1-\frac{w^2}{3!}+\frac{w^4}{5!}\mp \dots\right)=1$$

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wolfram alpha says it's 1 too but why? I think it should be 0. Thanks for the help. – gasko peter Mar 17 '13 at 18:23
@gaskopeter added – Dominic Michaelis Mar 17 '13 at 18:27
sin(w)/w = 0/0 no? why 1? – gasko peter Mar 17 '13 at 18:30
@gaskopeter added, $0/0$ is not $0$ the limit can be anything, here it is 1 – Dominic Michaelis Mar 17 '13 at 18:38
yeah exactly, ah other trivial example is having the limit $x\to 0$ of $\frac{2x}{x}$ this one is obviosly not $1$, so thats why we need further analisation – Dominic Michaelis Mar 17 '13 at 18:45

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