Question

Consider the function \[f(x,y)=\begin{cases}(x^2+y^2)\sin\dfrac{1}{x^2+y^2}, & (x,y)\neq(0,0)\\ 0, & (x,y)=(0,0)\end{cases}\] Which one of the following statements is correct for this function?

• A. \(f_x\) and \(f_y\) exist at (0,0), but are unbounded in any neighborhood of (0,0).
• B. f is not continuous at (0,0).
• C. f is continuous, but not differentiable at (0,0).
• D. f is continuous and differentiable at (0,0).

Hint:

Bound f(x,y) by r^2 for continuity, then check the differentiability limit directly using the squeeze theorem.

Answer 1

The Correct Option is D

Step 1: Continuity at the origin. Write \(r^2=x^2+y^2\). Since \(|\sin(1/r^2)|\le 1\), for \((x,y)\neq(0,0)\), \[|f(x,y)|=r^2\left|\sin\frac{1}{r^2}\right|\le r^2.\] As \((x,y)\to(0,0)\), \(r^2\to 0\), so \(f(x,y)\to 0=f(0,0)\). Hence f is continuous at (0,0).

Step 2: Partial derivatives at the origin. Using the definition, \[f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{h^2\sin(1/h^2)}{h}=\lim_{h\to 0} h\sin\frac{1}{h^2}=0,\] since \(|h\sin(1/h^2)|\le|h|\to 0\). By symmetry, \(f_y(0,0)=0\) as well. Both partial derivatives exist at (0,0) and equal 0.

Step 3: Differentiability at the origin. f is differentiable at (0,0) if \[L=\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-f_x(0,0)x-f_y(0,0)y}{\sqrt{x^2+y^2}}=0.\] This reduces to \[L=\lim_{r\to 0}\frac{r^2\sin(1/r^2)}{r}=\lim_{r\to 0} r\sin\frac{1}{r^2}=0,\] again by the squeeze theorem since \(|r\sin(1/r^2)|\le r\). So the limit is 0 and f is differentiable at (0,0), with total derivative equal to 0.

Step 4: Conclusion. f is both continuous and differentiable at (0,0). (For \((x,y)\neq(0,0)\) the partial derivative functions are unbounded near the origin, so f fails to be continuously differentiable there; continuity of the partials is only a sufficient condition for differentiability, not a necessary one, and Step 3 confirms differentiability directly.) \[\boxed{\text{f is continuous and differentiable at }(0,0)}\]