Function Definition and Analysis at (0,0)

The function given is defined as:

\[ f(x, y)=\left\{\begin{array}{l} \frac{x^2-y^2}{x^2+y^2} \text { for }(x, y) \neq(0,0) \\ 0 \text { for }(x, y)=(0,0) \end{array} \right. \]

We need to analyze several properties of this function at the point \((0,0)\).

Boundedness of the Function

Let's first examine the boundedness of the function \(f(x,y)\). For any point \((x,y) \neq (0,0)\), the value of the function is given by \(f(x,y) = \frac{x^2-y^2}{x^2+y^2}\).

We can consider the absolute value of the function:

\[ |f(x,y)| = \left|\frac{x^2-y^2}{x^2+y^2}\right| \]

We know that for any real numbers \(x\) and \(y\):

• \(x^2 \le x^2+y^2\)
• \(y^2 \le x^2+y^2\)

From these inequalities, we can see that \(x^2-y^2 \le x^2 \le x^2+y^2\), which means \(x^2-y^2 \le x^2+y^2\). Also, \(y^2-x^2 \le y^2 \le x^2+y^2\), which implies \(x^2-y^2 \ge -(x^2+y^2)\).

Combining these, we have:

\[ -(x^2+y^2) \le x^2-y^2 \le x^2+y^2 \]

For \((x,y) \neq (0,0)\), \(x^2+y^2 > 0\). Dividing the inequality by \(x^2+y^2\), we get:

\[ -1 \le \frac{x^2-y^2}{x^2+y^2} \le 1 \]

This means \(|f(x,y)| \le 1\) for all \((x,y) \neq (0,0)\). At \((0,0)\), \(f(0,0) = 0\), and \(|0| \le 1\). Therefore, for all \((x,y) \in \mathbb{R}^2\), \(|f(x,y)| \le 1\).

A function is bounded in a neighbourhood of a point if there exists a neighbourhood around that point where the function's values are limited between two numbers. Since \(|f(x,y)| \le 1\) for all \((x,y)\), the function is bounded everywhere, and thus it is certainly bounded in any neighbourhood of \((0,0)\).

So, the statement "f is bounded in a neighbourhood of (0, 0)" is true.

Continuity at (0,0)

For a function to be continuous at \((0,0)\), the limit as \((x,y) \to (0,0)\) must exist and be equal to \(f(0,0) = 0\). Let's examine the limit along different paths approaching \((0,0)\).

• Along the x-axis (y=0, \(x \neq 0\)):
  \( \lim_{(x,y) \to (0,0) \text{ along } y=0} f(x,y) = \lim_{x \to 0} f(x,0) = \lim_{x \to 0} \frac{x^2 - 0^2}{x^2 + 0^2} = \lim_{x \to 0} \frac{x^2}{x^2} = \lim_{x \to 0} 1 = 1 \)
• Along the y-axis (x=0, \(y \neq 0\)):
  \( \lim_{(x,y) \to (0,0) \text{ along } x=0} f(x,y) = \lim_{y \to 0} f(0,y) = \lim_{y \to 0} \frac{0^2 - y^2}{0^2 + y^2} = \lim_{y \to 0} \frac{-y^2}{y^2} = \lim_{y \to 0} -1 = -1 \)

Since the limits along different paths are different (1 and -1), the limit of \(f(x,y)\) as \((x,y) \to (0,0)\) does not exist. Therefore, the function \(f\) is not continuous at \((0,0)\).

So, the statement "f is continuous at (0, 0)" is false.

Consequently, the statement "f is not bounded in any neighbourhood of (0, 0)" must also be false, as we have already shown it is bounded.

Directional Derivatives at (0,0)

The directional derivative of \(f\) at \((0,0)\) in the direction of a non-zero vector \(\mathbf{v} = (v_x, v_y)\) is given by the limit:

\[ D_\mathbf{v} f(0,0) = \lim_{t \to 0} \frac{f((0,0) + t(v_x, v_y)) - f(0,0)}{t} = \lim_{t \to 0} \frac{f(tv_x, tv_y) - f(0,0)}{t} \]

For \(t \neq 0\), \((tv_x, tv_y) \neq (0,0)\) if \((v_x, v_y) \neq (0,0)\). In this case:

\[ f(tv_x, tv_y) = \frac{(tv_x)^2 - (tv_y)^2}{(tv_x)^2 + (tv_y)^2} = \frac{t^2 v_x^2 - t^2 v_y^2}{t^2 v_x^2 + t^2 v_y^2} = \frac{t^2(v_x^2 - v_y^2)}{t^2(v_x^2 + v_y^2)} = \frac{v_x^2 - v_y^2}{v_x^2 + v_y^2} \]

We have \(f(0,0) = 0\). So the limit becomes:

\[ D_\mathbf{v} f(0,0) = \lim_{t \to 0} \frac{\frac{v_x^2 - v_y^2}{v_x^2 + v_y^2} - 0}{t} = \lim_{t \to 0} \frac{1}{t} \left(\frac{v_x^2 - v_y^2}{v_x^2 + v_y^2}\right) \]

For this limit to exist and be finite, the term \(\frac{v_x^2 - v_y^2}{v_x^2 + v_y^2}\) must be equal to 0. This happens if and only if \(v_x^2 - v_y^2 = 0\), which means \(v_x^2 = v_y^2\), or \(|v_x| = |v_y|\). This condition is met only for directions along the lines \(y=x\) or \(y=-x\).

For any direction where \(|v_x| \neq |v_y|\) (e.g., the direction \((1,0)\) along the x-axis where \(v_x=1, v_y=0\), so \(v_x^2-v_y^2 = 1 \neq 0\)), the limit does not exist (it goes to \(\pm \infty\)).

Since the directional derivative does not exist for all possible directions, the statement "f has all directional derivatives at (0, 0)" is false.

Conclusion on Statements

Based on our analysis:

• f is continuous at (0, 0) - False
• f is bounded in a neighbourhood of (0, 0) - True
• f is not bounded in any neighbourhood of (0, 0) - False
• f has all directional derivatives at (0, 0) - False

The only true statement among the options provided is that f is bounded in a neighbourhood of (0, 0).