Maybe it's trivial, but I need certaintes about this. Consider
$$f(x, y) = \begin{cases} \dfrac{2xy^2}{x^2+y^2} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0) \end{cases}$$
I showed that the function is not continuous at the origin. Indeed by restricting to the straight lines though the origin $y = mx$ I get $0$ as a limit ($x\to 0)$.
Instead, taking the sequences $(x_n, y_n) = \left(\frac{1}{n^2}, \frac{1}{n}\right)$, I observed that
$$\lim_{n\to +\infty} f(x_n, y_n) = \frac{2}{3}$$
Hence the statement.
Besides this, can I still talk about existence of the derivatives? I'm asking because I managed to show that by using the definition:
$$f'_x(0, 0) = 0$$ $$f'_y(0, 0) = 0$$
Hence the derivatives exist everywhere, but they are not continuous since
$$\lim_{(x, y) \to (0, 0)} f'_x (x, y) = -2m^2$$
when restricting to $y = mx$ again, hence the limit does not exist. Similarly for $f'_y$.
After that, I also showed that $f(x, y)$ is not differentiable at $(0, 0)$, indeed
$$\lim_{(h, k)\to (0, 0)} \frac{f(h, k) - f(0, 0) - h\cdot f'_x(0, 0) - k\cdot f'_y(0, 0)}{\sqrt{h^2 +k^2}} = \sqrt{2}$$
Question: is this all right? Did I make any error?
Can I conclude that even if $f$ is not continuous at a point, its partial derivatives (along the axis) could exist (everywhere)?
$f(x, y) = \begin{cases} \dfrac{2xy^2}{x^2+y^2} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0) \end{cases}$
is continuous everywhere.
Why this contradict that the limit doesn't exist?
If the limit exists then it would be independent of path.To prove non existence of a limit, we choose a path(cleverly) such that the limit depends on the parameter because for two distinct values of parameter we get two different limit.
For an example : $f(x, y) = \begin{cases} \dfrac{2y^2}{x^2+y^2} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0) \end{cases}$
Now choose the path $y=mx$
Then $\lim_{x\to 0}f(x, mx) =\frac{m^2}{1+m^2}$
So along two different path $y=x$ and $y=2x$ we get two different value, so the limit at the origin doesn't exist.
Note: In case of your example, you haven't proved the dependency of limit on the parameter.
Again wrong.
$f(x_n, y_n) =\frac{1}{n^2+1}\to 0 =f(0,0)$
Both the above approach can't discard the discontinuity of $f$ at $(0, 0) $.
Infact $f$ is continuous at $(0, 0) $
$\begin{align}|f(x, y) -f(0, 0) |&=\left|\frac{2xy^2}{x^2+y^2}\right|\\&\le 2|x|\\&\le 2\sqrt{x^2+y^2}\end{align}$
Hence $\forall \epsilon>0$ , choose $\delta \in(0, \frac{\epsilon}{2}) $ such that $\forall (x, y) \in N_{\delta}((0,0))$ we have $|f(x, y) -f(0, 0) |<\epsilon$
Are the partial derivatives exist at $(0,0)$?
We can prove more than that "directional derivatives of $f$ at $(0, 0) $ along the direction of every unit vector exists and $0$"
Proof:
Let $u=(u_1, u_2) $ be a unit vector i.e $\|u\|=\sqrt{a^2+b^2}=1$
Then $\begin{align}D_{u}f(0, 0) &=\lim_{t\to 0}\frac{ f(tu_1, tu_2) -f(0, 0) }{t}\\&=\lim_{t\to 0} \frac{2tuv^2}{u^2+v^2}\\&=0\end{align}$
Let $e_1=(1, 0) $ and $e_2=(0,1)$ , then $\frac{\partial{f}}{\partial{x}}=D_{e_1}f(0, 0)=0$ and $\frac{\partial{f}}{\partial{y}}=D_{e_2}f(0, 0)=0$
This is again false. Existence of partial derivatives (even directional derivatives along every unit vector) at a point doesn't imply differentiability at that point.
Your example is good to enough to emphasize that fact.
Is $f$ differentiable at $(0, 0) $?
Ans: No
Proof: $\begin{align}&\frac{f(h,k)-f(0,0)-h\frac{\partial{f}}{\partial{x}}(0,0)-k\frac{\partial{f}}{\partial{y}}(0,0)}{\sqrt{h^2+k^2}}\\&=\frac{2hk^2}{(h^2+k^2)^{\frac{3}{2}}}\end{align}$
$\lim_{(h, k) \to (0, 0)} \frac{2hk^2}{(h^2+k^2)^{\frac{3}{2}}}$ doesn't exists.
Choose the path $k=mh$.