Discussion
Let \(f: \mathbb{R}^{2} \to \mathbb{R}\) be the function defined by
\[ f(x,y) = \begin{cases} \frac{5x^{2}y - 4y^{3}}{x^{2} + y^{2}}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases} \]
Which ONE of the following statements is FALSE?
Answer: D
The FALSE statement is: \(\frac{\partial f}{\partial y}\) is continuous at \((0,0)\).
At the origin, \(f(0,y) = \frac{-4y^{3}}{y^{2}} = -4y\), so \(\frac{\partial f}{\partial y}(0,0) = -4\). At a point \((x,0)\) with \(x \neq 0\), the linear term in \(y\) is \(\frac{5x^{2}y}{x^{2}} = 5y\), so \(\frac{\partial f}{\partial y}(x,0) = 5\). Approaching the origin along the \(x\)-axis the partial derivative stays at 5, not \(-4\), so it is not continuous there.
The other statements are true. Continuity: \(|f(x,y)| \leq \frac{5x^{2}|y| + 4|y|^{3}}{x^{2} + y^{2}} \leq 5|y| \to 0\). For \(\frac{\partial f}{\partial x}\) at \((0,1)\), \(f(x,1) = \frac{5x^{2} - 4}{x^{2} + 1}\) is even in \(x\), so its derivative at \(x = 0\) is 0. For \(\frac{\partial f}{\partial y}\) at \((1,0)\), \(f(1,y) = \frac{5y - 4y^{3}}{1 + y^{2}}\) has derivative 5 at \(y = 0\), the same value found above.
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