Let $f:\mathbb{R^2}\to\mathbb{R}$ be defined by $f(x,y) =
\begin{cases}
\dfrac{x^2y}{x^4+y^2}, & \text{if $(x,y)\neq(0,0)$}, \\
0, & \text{if $(x,y)=(0,0)$}
\end{cases}$
Which of the following statements is true regarding the continuity and existence of partial derivatives at $(0,0)$?
a) Both partial derivatives of $f$ exist at $(0,0)$ and $f$ is continuous at $(0,0)$
b) Both partial derivatives of $f$ exist at $(0,0)$ and $f$ is NOT continuous at $(0,0)$
c) One partial derivatives of $f$ does not exist at $(0,0)$ and $f$ is continuous at $(0,0)$
d) One partial derivatives of $f$ does not exist at $(0,0)$ and $f$ is NOT continuous at $(0,0)$
From what I worked out, $f$ is continuous at $(0,0)$. For the partial derivative part
$\dfrac{\partial f}{\partial x}=\dfrac{2xy(x^4+y^2)-x^2y(4x^3)}{(x^4+y^2)^2}$. Putting $y = mx$, I end up with $\dfrac{2m(-x^2+m^2)}{(x^2+m^2)^2}$. This depends on the value of m and hence, the the limit at $(0,0)$ should not exist.
I get a similar expression for the partial derivatives w.r.t. y, which will depend on m and hence the limit at $(0,0)$ should not exist . I am stuck now.
Please help.
Best Answer
First we determine whether or not $f$ is continuous at $(0,0)$. If we approach $(0,0)$ on the direction $x=y$, we have $$\lim_{x\to 0,y\to 0, x=y}f(x,y)\lim_{x\to 0}\frac{x^3}{x^4+x^2}=\lim_{x\to 0}\frac{1}{x+x^{-1}}=0$$ If we approach on the direction $y=x^2$, then $$\lim_{x\to 0,y\to 0, y=x^2}f(x,y)=\lim_{x\to 0}\frac{x^2\cdot x^2}{x^4+(x^2)^2}=\frac{1}{2}$$ Thus we conclude that $f$ is not continuous at $(0,0)$.
For the partial derivative. To determine whether $f_x(0,0)$ exists, simply apply the definition of partial derivative:
$$\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x}=0$$
Thus $f_x(0,0)=0$
Similarly
$$\lim_{y\to 0}\frac{f(0,y)-f(0,0)}{y}=0$$ thus $f_y(0,0)=0$