Examine the continuity of f : R2 → R at (0, 0), where for all (x, y) ∈ R2,f(x, y) :=





sin x sin

1

y

!

if y ̸= (0, 0),

0 if y = 0.

We are asked to examine the continuity of the function \( f: \mathbb{R}^2 \to \mathbb{R} \) at \( (0, 0) \), where \( f(x, y) \) is defined as follows:

\[
f(x, y) =
\begin{cases}
\sin(x) \sin\left(\frac{1}{y}\right), & \text{if } y \neq 0, \\
0, & \text{if } y = 0.
\end{cases}
\]

### Definition of Continuity:
A function \( f(x, y) \) is continuous at \( (0, 0) \) if:
\[
\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0).
\]
In this case, \( f(0, 0) = 0 \), so we need to check whether the limit of \( f(x, y) \) as \( (x, y) \to (0, 0) \) is 0.

### Step 1: Check the behavior along different paths.
To investigate whether the limit exists, we'll check the limit along several paths as \( (x, y) \to (0, 0) \).

#### Path 1: \( y = 0 \)
When \( y = 0 \), the function simplifies to:
\[
f(x, 0) = 0.
\]
Thus,
\[
\lim_{x \to 0} f(x, 0) = 0.
\]

#### Path 2: \( x = 0 \)
When \( x = 0 \), the function simplifies to:
\[
f(0, y) = 0 \quad \text{for all } y \neq 0.
\]
Thus,
\[
\lim_{y \to 0} f(0, y) = 0.
\]

#### Path 3: \( y = x \)
When \( y = x \), the function becomes:
\[
f(x, x) = \sin(x) \sin\left(\frac{1}{x}\right) \quad \text{for } x \neq 0.
\]
Since \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 as \( x \to 0 \), the product \( \sin(x) \sin\left(\frac{1}{x}\right) \) oscillates as well. Hence, the limit along this path does not exist.

### Step 2: Investigate general behavior as \( (x, y) \to (0, 0) \)
Consider the absolute value of the function for general \( (x, y) \neq (0, 0) \):
\[
|f(x, y)| = |\sin(x) \sin\left(\frac{1}{y}\right)|.
\]
Since \( |\sin(x)| \leq 1 \) and \( |\sin\left(\frac{1}{y}\right)| \leq 1 \), we have:
\[
|f(x, y)| \leq 1 \quad \text{for all } (x, y).
\]
However, as we have seen from the path \( y = x \), the function oscillates, and the limit does not settle to a single value as \( (x, y) \to (0, 0) \).

### Step 3: Conclusion
The function does not have a limit at \( (0, 0) \) along all possible paths, particularly due to the oscillatory behavior along certain paths (e.g., \( y = x \)). Therefore, the function is **not continuous** at \( (0, 0) \).

Would you like further details or have any questions?

Here are 5 questions related to this problem:
1. How would the behavior of the function change if we altered the definition for \( f(x, y) \) when \( y = 0 \)?
2. Can you provide another example of a function that is discontinuous at a specific point due to oscillation?
3. What does it mean for a function to be continuous on the entire domain \( \mathbb{R}^2 \)?
4. How do the paths you choose affect the evaluation of limits in two-variable functions?
5. How would you apply the epsilon-delta definition of continuity to this problem?

**Tip:** When evaluating limits for multivariable functions, always check behavior along multiple paths to ensure consistency of the limit.

Examine the continuity of f : R2 → R at (0, 0), where for all (x, y) ∈ R2,f(x, y) := sin(x) * sin(1/y) if y ≠ 0, and 0 if y = 0.

Explore the continuity of a multivariable function f(x, y) defined by sin(x) * sin(1/y) for y ≠ 0, and 0 if y = 0, at the point (0, 0). This problem includes step-by-step analysis using different paths and discusses oscillatory behavior. Ideal for undergraduate students studying multivariable calculus.

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