Check the continuity and the differentiability of a function of two variables

Question

Consider a function $f: R^2 \rightarrow R$ defined by $f(x,y)=\frac {xy(x+y)} {x^2+y^2}$ if $(x,y)\ne(0,0)$ and $f(x,y)=0$ if $(x,y)=(0,0)$. The first question asks about the continuity of $f$ in $(0,0)$ and the second question asks about the differentiability of $f$ in $(0,0)$. I think I did the questions well, but I'm not completely sure. For continuity $lim_{x\rightarrow0,y\rightarrow0}\frac {xy(x+y)} {x^2+y^2}$ can be computed using polar coordinates: $lim_{r\rightarrow0}\frac {r^2cos\theta sin\theta r(cos\theta+sin\theta)} {r^2}=0$ by the r in the numerator. So $f$ is continuos in $(0,0)$. For differentiability we have to consider the limit $\lim_{h\rightarrow0,k\rightarrow0}\frac {f(x_0+h,y_o+k)-f(x_0,y_o)-Df[x_0,y_0](h,k)} {\sqrt{h^2+k^2}}$ where $(x_0,y_0)=(0,0)$. Both the partial derivatives of $f(x,y)$ in $(0,0)$ are $0$ so $Df[x_0,y_0](h,k)=0$. So the limit is simply: $\lim_{h\rightarrow0,k\rightarrow0}\frac {hk(h+k)} {(h^2+k^2)^\frac {3} {2}}$ and by using the y-axis $x=0$ this limit is $0$ while by using $y=x$ this limit is $\frac {1} {\sqrt{2}}$ so this limit does not exists. This tells us that $f$ is not differentiable in $(0,0)$.

Answer 1

Rigorously proving continuity at $(0, 0)$:

$$\bigg|\frac{xy(x+y)}{x^2+y^2}\bigg| \leq \frac{\vert x\vert \vert y \vert (\vert x+y\vert)}{x^2+y^2} \leq \frac{(x^2+y^2)(\vert x+y\vert)}{x^2+y^2} = \vert x+y \vert\to 0\quad \text{as} (x, y) \to (0, 0) $$

Hence $f(x, y)$ is continuous at the origin.

Also consider that to check for the values of partial derivatives you don't have to calculate $\frac{\partial f}{\partial x}$ and then evaluate it at $(0, 0)$ (or take the limit): you need to prove the existence through the definition (which you may have actually done, but it's unclear from your post).

$$f'_x(0, 0) = \lim_{h\to 0} \frac{f(0+h, 0) - f(0, 0)}{h} = \lim_{h\to 0} \frac{f(h, 0)}{h} = 0$$

The same for $f'y$.

Finally, beware about the differentiation for you have

$$\lim_{(x, y) \to (0, 0)} \frac{xy(x+y)}{\sqrt{x^2+y^2}(x^2+y^2)}$$

And when choosing the restricition $x = y$ the limit is not really $\frac{1}{\sqrt{2}}$ but it depends on the direction:

$$\lim_{x\to 0^+} \frac{x}{\sqrt{2} \sqrt{x^2}} = \frac{1}{\sqrt{2}}$$

$$\lim_{x\to 0^-} \frac{x}{\sqrt{2} \sqrt{x^2}} = -\frac{1}{\sqrt{2}}$$

However one limit of that kind is just enough to prove $f$ is not differentiable.