Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$
I need to specify the value of $f(x,y)$ which makes the given function continuous.
I had tried to give a arbitral relationship between $y$ and $x$, but as $(x, y)$ gets closer to $(0, 0)$ the arbitral movement cannot always be represented as a functional representation between $y$ and $x$.
How could I solve this problem rigorously?
Recalling that $|x|\,|\cos(x)|\le |\sin(x)|\le |x|$ for $0\le |x|\le \pi/2$, we assert that for $x^2+y^2\le \pi/2$
$$|\cos(x^2+y^2)|\le \left|\frac{\sin(x^2+y^2)}{x^2+y^2}\right|\le 1$$
whence application of the squeeze theorem reveals
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1$$
I want to add something might be too long for a comment, that aside from showing the calculations, it's also important to know what we are trying to do here.
In order to rigorously prove it is continuous at $(0,0)$, you cannot just take paths approaching it, but you need to use definition of continuity at a point, i.e.
$$\forall \epsilon >0, \exists r > 0, \text{ s.t. whenever } (x,y) \in B_r((0,0)), \text { we have } |f(x,y) -1 | < \epsilon$$
And Mark Viola's technique could be used to show the above is true (+1).
For $\lim_{(x,y)\rightarrow (0,0)} \frac{sin(x^2 +y^2}{x^2+y^2}$, using $r^2=x^2+y^2$ we get $\lim_{r\rightarrow 0} \frac{sin(r^2)}{r^2}$=$\lim_{r \rightarrow 0} \frac{2 \cdot r \cdot \cos(r^2)}{2r}$=$\lim_{r \rightarrow 0} \cos(r^2)=1$, therefore $\lim_{(x,y)\rightarrow (0,0)} \frac{sin(x^2 +y^2}{x^2+y^2}=1$.
