Differentiability question in multivariable function

Question

Let:

$f(x) = (x^2 + y^2)\sin(\dfrac{1}{x^2+y^2})$ , if $(x,y) \neq (0,0) $ and
$f(x) = 0$ , if $(x,y) = (0,0)$

Show that $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ exists everywhere, but are discontinuous in $(0,0)$. Also show that $f(x)$ is differentiable in $(0,0)$.

Hint: Observe that the differentiability problem is in the point $(0,0)$ and verifiy that $\dfrac{\partial f}{\partial x}(0,0) = 0 = \dfrac{\partial f}{\partial y}(0,0)$ and, in this case, $\nabla f(0,0) = (0,0)$. And to show that $f$ is differentiable in (0,0), you only need to show that: $$\lim_{\sqrt{x^2 + y^2} \to \infty} \dfrac{f(x,y)-f(0,0)}{\sqrt{x^2 + y^2}} = 0$$

I managed to prove the equality of the limit given in the hint, but i can't see why proving that limit does prove that the function is differentiable in $(0,0)$. Can anyone explain that to me?

I also managed to answer the other questions of the problem, i'm just stuck in the differentiability on $(0,0)$.

Thanks!

Answer 1

It must be a mistake, but the limit $$\lim_{(x,y)\to(0,0)}\frac{|f(x,y)-f(0,0)|}{\|(x,y)\|}=0$$ it sufficient to show that $f$ is differentiable at $(0,0)$. This follows from the definition:

A function $f:\mathbb R^n\to\mathbb R^m$ is differentiable at $x_0\in\mathbb R^n$ if there exists a linear transformation $\textsf T:\mathbb R^n\to\mathbb R^m$ such that $$\lim_{x\to x_0}\frac{\|f(x)-f(x_0)-\textsf{T}(x-x_0)\|}{\|x-x_0\|}=0$$

In this case, the linear transformation $\nabla f(0,0):\mathbb R^2\to\mathbb R$ is given by $$\nabla f(0,0)(x,y)=(0,0)\cdot(x,y)=0$$