Answer
See explanation
Work Step by Step
To analyze the continuity of the function
\[
f(x, y) = \begin{cases}
\frac{\sin(xy)}{xy} & \text{if } xy \neq 0 \\
1 & \text{if } xy = 0
\end{cases}
\]
we need to understand how \( f(x, y) \) behaves as \((x, y) \to (0, 0)\) and whether the limit matches the value defined at the origin, \( f(0, 0) = 1 \).
### Step 1: Find the Limit of \( f(x, y) \) as \( (x, y) \to (0, 0) \)
Let’s examine the behavior of \( \frac{\sin(xy)}{xy} \) as \((x, y) \to (0, 0)\).
Recall that for any variable \( u \), \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \). Here, let \( u = xy \), so as \( (x, y) \to (0, 0) \), we have \( u \to 0 \). Therefore,
\[
\lim_{(x, y) \to (0, 0)} \frac{\sin(xy)}{xy} = 1
\]
This suggests that as \((x, y) \to (0, 0)\), \( f(x, y) \) approaches \(1\), which is the value of \(f(0, 0)\).
### Step 2: Continuity of \( f(x, y) \) at \( (0, 0) \)
To be continuous at \((0, 0)\), we need:
1. \(f(0, 0) = 1\)
2. \(\lim_{(x, y) \to (0, 0)} f(x, y) = 1\)
Since both of these conditions are satisfied, \(f(x, y)\) is continuous at \((0, 0)\).
