f(x,y) = \begin{cases} \frac{\sin(x+y)}{x+y}, & \text{if } x + y \neq 0 \\\lim_{t \to 1} \frac{\sin t}{t}, & \text{if } x + y = 0 \end{cases}
1. For points where x + y \neq 0, the function is given by f(x,y) = \frac{\sin(x+y)}{x+y}.
2. For points where x + y = 0, we evaluate the limit:
\lim_{t \to 1} \frac{\sin t}{t} = \frac{\sin 1}{1}.
Therefore, the continuous function f(x,y) is
f(x,y) = \begin{cases} \frac{\sin(x+y)}{x+y}, & \text{if } x + y \neq 0 \\\frac{\sin 1}{1}, & \text{if } x + y = 0 \end{cases}