Let $f:\Bbb R^2 \rightarrow \Bbb R $ be defined by
$$f(x)= \begin{cases}
f(x,y)=\dfrac{2x^3+3y^3}{x^2+y^2}\ \ \ &(x,y)\ne(0,0) \\
0 & (x,y) = (0,0) \\
\end{cases}
$$
Let $f_x(0,0)$ and $f_y(0,0)$ denote the first order partial derivative of $f(x,y)$ with respect to $x$ and $y$ respectively,at the point $(0,0)$. The which one of the following statements is TRUE?
$(A)$ $f$ is continuous at $(0,0)$ but $f_x(0,0)$ and $f_y(0,0)$ do not exist.
$(B)$ $f$ is differentiable at $(0,0)$
$(C)$ $f$ is not differentiable at $(0,0)$
$(D)$ $f$ is not continuous at $(0,0)$ but $f_x(0,0)$ and $f_y(0,0)$ exist.
My attempt:
I was checking continuity so:
Choosing path $y=mx$
$\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3y^3}{x^2+y^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3m^3x^3}{x^2+m^2x^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3m^3x^3}{x^2+m^2x^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{x(2+3m^3)}{1+m^2}=0$
$f(0,0)=0=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3y^3}{x^2+y^2}$
Makes it continuous so option $D$ cancels out
Checking partial differentiability:
$f_x(0,0)=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3y^3}{x^2+y^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{(x^2+y^2)\cdot(6x^2)-(2x^3+3y^3) \cdot (4x)}{(x^2+y^2)^2}$
Choosing path $y=mx$
$\lim_{(x,y)\rightarrow(0,0)}\dfrac{(x^2+m^2x^2)\cdot(6x^2)-(2x^3+3m^3x^3) \cdot (4x)}{(x^2+m^2x^2)^2}$
$\lim_{(x,y)\rightarrow(0,0)}\dfrac{(1+m^2)\cdot(6)-(2+3m^3) \cdot (4)}{(1+m^2)^2}$
For different values of $m$ we will have different answers so the limit is not unique.
$A$ option cancels out
So a function to be differentiable it must satisfy the following:
$$\lim_{(h,k)\rightarrow(0,0)}=\dfrac{f((0+h),(0+k))-f(x,y)-h \cdot f_x(0,0)-k \cdot f_y(0,0)))}{\sqrt{h^2+k^2}}$$
but $f_x$ and $f_y$ doesn't exist so its not differentiable.
Am I thinking it right so only option left is $C$
Best Answer
For the first part we have that
$$\frac{|2x^3+3y^3|}{x^2+y^2} \leq \frac{2|x^3|}{x^2+y^2}+\frac{3|y^3|}{x^2+y^2} \leq 2|x|+3|y|$$
thus the limit goes to $0$ by squeeze theorem and the function is continuous. For differentiability we must compute the partials at $(0,0)$ by definition:
$$f_x(0,0) = \lim_{x\to0} \frac{f(x,0)-f(0,0)}{x} = \lim_{x\to0} \frac{2x-0}{x} = 2$$
and similarly $f_y(0,0) = 3$. Therefore we need to check whether the following limit exists and equals $0$:
$$\lim_{(x,y)\to0} \frac{f(x,y)-f(0,0)-2x-3y}{\sqrt{x^2+y^2}}$$
I'll leave this last step for you to check.