Discussion
Let \(f: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) and \(g: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) be two bounded and differentiable functions. Define \(F: \mathbb{R}^{2} \to \mathbb{R}\) by
\[ F(x, y) = \begin{cases} x f(y) + y^{2} g(x), & xy \neq 0, \\ 0, & xy = 0. \end{cases} \]
Consider the following statements.
P: \(F\) is continuous at \((0, 0)\).
Q: Both the partial derivatives \(\frac{\partial F}{\partial x}\) and \(\frac{\partial F}{\partial y}\) exist at \((1, 0)\).
Then
Answer: A
P is correct and Q is NOT correct.
P: let \(|f| \leq M\) and \(|g| \leq M\). Then \(|F(x, y)| \leq M|x| + My^{2} \to 0\) as \((x, y) \to (0, 0)\), and \(F(0,0) = 0\), so \(F\) is continuous at the origin.
Q: on the line \(y = 0\), \(F(x, 0) = 0\) for all \(x\), so \(\frac{\partial F}{\partial x}(1, 0) = 0\) exists. But
\[ \begin{aligned} \frac{\partial F}{\partial y}(1, 0) &= \lim_{y \to 0} \frac{F(1, y) - 0}{y} \\ &= \lim_{y \to 0} \frac{f(y) + y^{2}g(1)}{y} \end{aligned} \]
which exists only if \(\frac{f(y)}{y}\) has a limit. A bounded differentiable \(f\) on \(\mathbb{R} \setminus \{0\}\) need not tend to 0 at 0; with \(f(y) = 1\) the quotient \(\frac{1}{y}\) blows up. So both partial derivatives need not exist, and Q fails.
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