# Proving differentiability, continuity and partial derivatives of the following two variables function

Prove the existence or not existence of continuity, differentiability and the continuity of partial derivatives of the following function two variables at $$(0,0)$$

$$f(x,y) = \begin{cases} (x-y)^{2}sin \frac{1}{x^{2}-y}, & \mbox{if } (x,y) \neq (0,0) , \\ 0, & \mbox{if } (x,y) =(0,0)\end{cases}$$

At first I wanted to prove this function was not continuous at $$(0,0)$$ therefore I would prove this function is not differentiable at $$(0,0)$$, as well. I tried to compute the limit of the function as we aproach to $$(0,0)$$ with two different trajectories,lets say $$y=x$$ and $$y=2x$$ but this doesnt seems to work. So maybe this function is continuous and differentiable. But I cant see the $$\epsilon- \delta$$ trick here. In order to prove differentiability I need to compute partial derivatives and show this partial derivatives are continuous isnt? But its seems harder to prove continuity in the partia derivatives since I cannot even compute continuity over the original function. :(

As $$\frac{\partial f}{ \partial x}(0,0)=lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h}= lim_{h \to 0} \frac{f(h,0)}{h}=lim_{h \to 0} \frac{(h)^{2}sin \frac{1}{h^{2}}}{h}$$

but I dont know how to reduce this into something that gives me an insight about the differentiability or continuity of this partial derivative. In the same way I got

$$\frac{\partial f}{ \partial y}(0,0)=\frac{h^{2}sin \frac{1}{-h^{2}}}{h}.$$

• The function doesn't appear to be defined when $y = x^2$ (except when $x=y=0$). Is that intentional? Also, I'm assuming your $(x,y) = (0,0)$ and $(x,y) \neq (0,0)$ should be reversed.
– user169852
May 7, 2021 at 20:26
• You are absolutely right! Let me fix the typos @Bungo
– Sok
May 7, 2021 at 20:32
• For continuity of $f$, you can observe that $\left|(x-y)^2\sin\frac{1}{x^2-y}\right| \leq (x-y)^2$ since $|\sin \theta| \leq 1$ for any real $\theta$. And $(x-y)^2 = x^2 -2xy + y^2$ becomes arbitrarily close to zero as $(x,y) \to (0,0)$.
– user169852
May 7, 2021 at 20:42
• Note that $-2xy \leq x^2 + y^2$ (because if you add $2xy$ to both sides you get the equivalent $0 \leq x^2 + 2xy + y^2 = (x+y)^2$, which is obviously true). Therefore, $x^2 - 2xy + y^2 \leq x^2 + (x^2 + y^2) + y^2 = 2x^2 + 2y^2 = 2(x^2 + y^2)$. Taking square roots gives you the bound $\sqrt{x^2 - 2xy + y^2} \leq \sqrt{2} \sqrt{x^2 + y^2}$.
– user169852
May 7, 2021 at 22:21
• An alternative simpler argument: note that $(x,y) \to 0$ if and only if $x \to 0$ and $y \to 0$. And the latter two imply that $x^2 - 2xy + y^2 \to 0$ because of the rules involving sums and products of limits.
– user169852
May 7, 2021 at 22:24

According to Wikipedia $$f$$ is differentiable at the origin if there exists a linear map $$J:\mathbb{R}^2 \rightarrow \mathbb{R}$$ such that $$\lim_{h \rightarrow 0}\frac{|f(h_1,h_2)-f(0,0)-J(h_1,h_2)|}{\|(h_1,h_2)\|_{\mathbb{R}^2}}=0$$ Take $$J\equiv 0$$ on $$\mathbb{R}^2$$ so that $$\frac{|f(h_1,h_2)-f(0,0)-J(h_1,h_2)|}{\|(h_1,h_2)\|_{\mathbb{R}^2}}=\frac{(h_1-h_2)^2\Big|\sin\Big(\frac{1}{h_1^2-h_2}\Big)\Big|}{\sqrt{h_1^2+h_2^2}}\leq \frac{(h_1-h_2)^2}{\sqrt{h_1^2+h_2^2}}\longrightarrow 0$$ as $$(h_1,h_2)\longrightarrow 0$$. This proves that $$f$$ is differentiable at $$(0,0)$$.
Note: $$(h_1,h_2)$$ cannot tend to the origin along any arbitrary path. The manner in which we take $$(h_1,h_2)$$ to the origin must avoid intersecting the parabola $$h_2=h_1^2$$. Most textbooks I've seen allow for the consideration of such limits.
• @MathewPilling This was very helpful. The way I calculated the partial derivatives of this function at $(0,0)$ shows the derivatives are continuous at $(0,0)$? Can they be reducted and why the tend to $0$?
• You can show that $f_x(0,0)$ and $f_y(0,0)$ exist and are both equal to $0$ by using the definition of the partial derivative. If you want to show $f_x$ and $f_y$ are continuous at $(0,0)$, you need to show that $f_x(x,y),f_y(x,y)\longrightarrow 0$ as $(x,y)\longrightarrow (0,0)$. It's not enough to show the partial derivatives exist at $(0,0)$ to say that they're continuous at $(0,0)$ May 8, 2021 at 1:41
• Appreciate it! I will try to compute the limit of the partial derivatives as $(x,y) \to (0,0)$