Originally Posted by
Prove It I assume that you actually mean
$\displaystyle f(x, y) = \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}$ for $\displaystyle (x, y) \neq (0, 0)$
and $\displaystyle f(x, y) = 1$ for $\displaystyle (x, y) = (0, 0)$.
To show that this is continuous, you'd have to show that
$\displaystyle \lim_{(x, y) \to (0, 0)} f(x, y) = 1$.
$\displaystyle \lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}$
$\displaystyle = \lim_{r \to 0} \frac{\sin{r^2}}{r^2}$
since $\displaystyle r^2 = x^2 + y^2$.
It is well known that $\displaystyle \lim_{X \to 0} \frac{\sin{X}}{X} = 1$ (and can be shown using the sandwich theorem).
Therefore $\displaystyle \lim_{r \to 0} \frac{\sin{r^2}}{r^2} = 1$.
Since this limit is $\displaystyle 1$, the function is continuous.