I have the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $f(x,0)=1+x^2$, $f(0,y)=1+y^2$, and $f(x,y)=0$ if $x\neq0$ and $y\neq0$ and I need to answer the following questions
- Compute $\displaystyle\frac{\partial f}{\partial x}(0,0)$ and $\displaystyle\frac{\partial f}{\partial y}(0,0)$.
- Is $f$ continuons in $(0,0)$?
- Do $\displaystyle\frac{\partial f}{\partial x}(1,0)$ and $\displaystyle\frac{\partial f}{\partial x}(0,1)$ exist?
- What is the domain of $\displaystyle\frac{\partial f}{\partial x}$?
My progress
1)
$$\displaystyle\frac{\partial f}{\partial x}(0,0)=\lim_{x\rightarrow0}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\rightarrow0}\frac{1+x^2-(1+0^2)}{x}=0$$ and $$\displaystyle\frac{\partial f}{\partial y}(0,0)=\lim_{y\rightarrow0}\frac{f(0,y)-f(0,0)}{y-0}=\lim_{y\rightarrow0}\frac{1+y^2-(1+0^2)}{y}=0.$$
2) Let's consider these curves: $\gamma_1(t)=(t,0)$ and $\gamma_2(t)=(t,t)$. For the first, we have
$$\lim_{t\rightarrow0}f(\gamma_1(t))=1+0^2=1$$ and for the second $$\lim_{t\rightarrow0}f(\gamma_2(t))=\lim_{t\rightarrow0}0=0,$$ because $f$ is zero for $(x,y)\neq(0,0)$.
3) The first must be $$\displaystyle\frac{\partial f}{\partial x}(1,0)=\lim_{x\rightarrow1}\frac{f(x,0)-f(1,0)}{x-1}=\lim_{x\rightarrow1}\frac{1+x^2-(1+1^2)}{x-1}=\lim_{x\rightarrow1}(x+1)=2$$
and the second as follows
$$\displaystyle\frac{\partial f}{\partial x}(0,1)=\lim_{x\rightarrow0}\frac{f(x,1)-f(0,1)}{x-0}=\lim_{x\rightarrow0}\frac{0-(1+1^2)}{x}=-\lim_{x\rightarrow0}\frac{2}{x}$$ that does not exist
4) In the item (1) we see that $(0,0)$ belongs to $D=$ domain of $\displaystyle\frac{\partial f}{\partial x}$. For points like $(x_0,0)$, $x_0\neq0$ we have $$\displaystyle\frac{\partial f}{\partial x}(x_0,0)=\lim_{x\rightarrow0}\frac{f(x,0)-f(x_0,0)}{x-0}=\lim_{x\rightarrow x_0}\frac{1+x^2-(1+x_0^2)}{x}=\lim_{x\rightarrow x_0}\frac{x^2-x_0^2}{x-x_0}=2x_0$$
So the whole $x$ axis is in $D$. Finnaly, given $(0,y_0)$, $y_0\neq0$, we have $$\displaystyle\frac{\partial f}{\partial x}(0,y_0)=\lim_{x\rightarrow0}\frac{f(x,y)-f(0,y_0)}{x-0}=\lim_{x\rightarrow x_0}\frac{0-(1+y_0^2)}{x}=-\lim_{x\rightarrow x_0}\frac{1+y_0^2}{x}$$ which does not exist for every $y_0\in\mathbb{R}$
Therefore, $D=\left\{(x,y)\in\mathbb{R}\,|\,y\neq0)\right\}$.