Is This Function Continuous At (0,0)?

Hey guys, let's dive into a bit of a mind-bender in the world of calculus! We're going to explore a function that looks pretty innocent at first glance but has a sneaky little twist. Today's mission, should you choose to accept it, is to show that the function f(x, y) is not continuous at the point (0, 0). This function is defined as:

$$f(x, y) = \begin{cases} y \sin\left(\frac{1}{x}\right) + x \sin\left(\frac{1}{y}\right) & \text{if } x \neq 0 \text{ and } y \neq 0 \\ 1 & \text{otherwise} \end{cases}$$

Now, when we talk about continuity in multivariable calculus, especially at a specific point like (0, 0), we're basically asking if the function's value at that point matches what it should be if we approach it from any direction. For a function to be continuous at a point (a, b), three conditions must be met:

1. f(a, b) must be defined.
2. The limit of f(x, y) as (x, y) approaches (a, b) must exist.
3. The limit must equal the function's value at (a, b), i.e., $\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$.

In our case, the point of interest is (0, 0). Let's break down these conditions for our specific function, f(x, y). First off, condition 1 is satisfied. The function is explicitly defined for all points except when x = 0 or y = 0. At (0, 0), the function is defined as f(0, 0) = 1. So, we've got a defined value there. That's a good start, right?

Now for the real challenge: condition 2. Does the limit of f(x, y) exist as (x, y) approaches (0, 0)? This is where things get interesting. For a limit to exist at a point in multivariable calculus, the function must approach the same value no matter how we approach that point. Think of it like approaching your favorite cafe – you could come from the north, south, east, or west, and you should still end up at the same cafe door. If you end up at different places depending on your route, well, that's not a single cafe, is it? The same applies to limits.

To show that a limit does not exist, we often try to find two different paths of approach that lead to different limiting values. If we can find just one path that gives a different limit than another path, or if a path leads to an undefined limit, then the overall limit does not exist.

Let's consider approaching (0, 0) along the x-axis. Along the x-axis, y = 0. According to the definition of our function, when y = 0 (and x is not necessarily 0), f(x, y) = 1. So, as (x, y) approaches (0, 0) along the x-axis, the function value is always 1. This means the limit along the x-axis is 1:

$$\lim_{(x,y) \to (0,0) \text{ along } y=0} f(x, y) = \lim_{x \to 0} f(x, 0) = \lim_{x \to 0} 1 = 1$$

Similarly, let's approach (0, 0) along the y-axis. Along the y-axis, x = 0. Again, when x = 0 (and y is not necessarily 0), f(x, y) = 1. So, as (x, y) approaches (0, 0) along the y-axis, the function value is also always 1. This gives us a limit of 1 along the y-axis:

$$\lim_{(x,y) \to (0,0) \text{ along } x=0} f(x, y) = \lim_{y \to 0} f(0, y) = \lim_{y \to 0} 1 = 1$$

Okay, so far, two paths give us a limit of 1. This doesn't prove continuity yet, but it's what we'd expect if the function were continuous, since f(0, 0) = 1. The issue arises when we consider paths where neither x nor y is zero. This is where the y sin(1/x) + x sin(1/y) part of the function definition kicks in.

Let's try approaching (0, 0) along a path where x and y are related. A common strategy is to use a line y = mx or a curve like y = x^k. Let's try the line y = x. If we approach (0, 0) along the line y = x, then for any point on this line (except (0,0)), x ≠ 0 and y ≠ 0. So we use the first part of the function definition:

$$f(x, x) = x \sin\left(\frac{1}{x}\right) + x \sin\left(\frac{1}{x}\right) = 2x \sin\left(\frac{1}{x}\right)$$

Now, let's find the limit as (x, y) approaches (0, 0) along this line y = x. This is equivalent to finding the limit of $2x \sin(1/x)$ as $x \to 0$:

$$\lim_{(x,y) \to (0,0) \text{ along } y=x} f(x, y) = \lim_{x \to 0} 2x \sin\left(\frac{1}{x}\right)$$

To evaluate this limit, we need to remember a crucial fact about the sine function: $\sin(1/x)$ is always bounded between -1 and 1, no matter how close x gets to 0 (as long as $x \neq 0$). So, we have:

$$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$

Multiplying by $2x$ (and considering cases where $2x$ is positive or negative), we get:

$$-|2x| \leq 2x \sin\left(\frac{1}{x}\right) \leq |2x|$$

As $x \to 0$, both $-|2x|$ and $|2x|$ approach 0. By the Squeeze Theorem (or Sandwich Theorem), this means:

$$\lim_{x \to 0} 2x \sin\left(\frac{1}{x}\right) = 0$$

So, along the line y = x, the limit of f(x, y) as (x, y) approaches (0, 0) is 0.

Now we have a problem! We found one path (approaching along the axes) where the limit is 1, and another path (approaching along y = x) where the limit is 0. Since the limit as (x, y) approaches (0, 0) must be the same regardless of the path taken, and we've found two different results (1 and 0), this means the limit of f(x, y) as (x, y) approaches (0, 0) does not exist.

Because condition 2 for continuity (the limit must exist) is not met, the function f(x, y) is not continuous at the point (0, 0). It's a classic example showing that just because a function has a defined value at a point, and limits along some paths exist and match that value, it doesn't guarantee overall continuity. You've got to check all possible paths! Pretty wild, huh? This discontinuity arises because the terms $y \sin(1/x)$ and $x \sin(1/y)$ become arbitrarily small and approach zero as $(x,y) o (0,0)$ when $x eq 0$ and $y eq 0$, while the function is defined to be 1 at $(0,0)$. The