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