If $f(x,y)=9-x^2-y^2$ if $x^2+y^2\leq9$ and $f(x,y)=0$ if $x^2+y^2>9$ study what happens at $(3,0)$

Question

If$$f(x,y)=\begin{cases}9-x^2-y^2&\text{if }x^2+y^2\leq9\\0&\text{if }x^2+y^2>9\end{cases}$$study the continuity and existence of partial derivative with respect to $y$ at point $(3,0)$.

The graph of the domain of $f$ is:

Continuity study at $(3,0)$:

$f(3,0)=9-3^2-0=0$, but I do not know how to find $$\lim_{(x,y)\to(3,0)}f(x,y).$$ I tried the following: $$\lim_{(x,y)\to(3,0)}f(x,y)=\left\{\begin{array}{l}\displaystyle\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}(9-(x^2+y^2))=9-9=0\\\displaystyle\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}\text{??}=\text{??????}\end{array}\right.$$ but then I realized that the "curve" $C_2$ is actually NOT a curve but a set of infinite points, as shown in the previous image.

Existence of partial derivative with respect to $y$ at $(3,0)$:

I know that I need to study whether $$\frac\partial{\partial y}f(3,0)=f'((3,0);(0,1))=f_y(3,0)=\lim_{h\to0}\frac{f((3,0)+h(0,1))-f(3,0)}{h}=\lim_{h\to0}\frac{f(3,h)}{h}$$ exists or not, but I am not able to even find that limit.

Any help? Thanks!!

Answer 1

Regarding the partial derivative question, note that $f(3,y) = 0$ for all values of $y$. (The line $x=3$ touches the circle $x^2+y^2=9$ tangentially and is otherwise completely outside, where the function is identically $0$.) Thus, $\frac{\partial f}{\partial y}(3,0) = 0$.

Now, if $a^2+b^2>9$, $f$ is identically $0$ in a neighborhood of $(a,b)$ and so $\frac{\partial f}{\partial y}(a,b) = 0$. If $a^2+b^2<9$, then $\frac{\partial f}{\partial y}(a,b) = -2b$. What happens if $a^2+b^2=9$ but $b\ne 0$? Then $\frac{\partial f}{\partial y}(a,b)$ does not exist: \begin{multline*} \lim_{h\to 0+}\frac{f(a,b+h)-f(a,b)}h = 0, \quad\text{whereas}\\ \lim_{h\to 0-}\frac{f(a,b+h)-f(a,b)}h = -2b\ne 0. \end{multline*} Therefore, $\frac{\partial f}{\partial y}$ is not continuous at $(3,0)$, as it is not even defined at all points nearby. Perhaps the question wasn't asking about this; it's a bit ambiguous.

Answer 2

Continuty at (3,0) : When $|(x,y)-(3,0)|=\sqrt{(x-3)^2+y^2}<\delta<1$ for some $\delta>0$ then $|f(x,y)|\leq 9-x^2-y^2=(3-x)(3+x)\leq 6 (3-x)\leq 6\sqrt{(x-3)^2+y^2}<6\delta.$ This implies that $f$ is continuous at $(3,0).$

Partial differential wrt $y$ : $f(3,y)=0$ for any $y$ and hence $f$ has a partial derivative at $(3,0)$ wrt $y$ and the partial derivative is $0.$

Answer 3

We have the piece-wise defined surface. The two surfaces are continuous and if the limits are equal at the join, then the surface is continuous.

And since the two surfaces are continuous we can just plug $(3,0)$ in for each.

$\lim_\limits{(x,y)\to (3,0)} 9-x^2-y^2 = 0$

$\lim_\limits{(x,y)\to (3,0)} 0 = 0$

Another way to do it would be to say:

$\forall \epsilon >0, \exists \delta > 0 : d((x,y),(3,0))< \delta \implies |f(x,y)|<\epsilon$

$d((x,y),(3,0))$ is the distance between $(x,y)$ and $(3,0)$

And what distance metric do we want to use? How about the taxicab metric?

Then the maximal distance some point $x_1,y_1$ can be from the point $(3,0)$ would be $(3+\pm \delta, \pm \delta)$

$f(x,y) < |6\delta| + |2\delta^2|$

Lets demand that $\delta < 1$ then $\delta^2 < \delta$

$\delta = \min (\frac {\epsilon}{8},1)$