Problem: Consider the function $f: \mathbb{R} ^2 \to \mathbb{R} ^2$ defined by
\[
f(x,y)=
\begin{cases}
1, &\text{ if } xy = 0 ;\\
0, &\text{ if } \text{ otherwise}.
\end{cases}
\]
Show that the partial derivatives of $f$ at $(0,0)$ exist but the function is not continuous at $(0,0)$.
Use the definition of partial derivatives at a point. For example,
\[
\frac{\partial f}{\partial x} (a,b) = \lim_{h \to 0} \frac{f(a+h,b) - f(a,b)}{h}
\]
A function $f$ is not continuous at a point $x$ if there exists a sequence $x_n$ converging to $x$ but $f(x_n)$ does not converge to $f(x)$. Consider the sequence $\left( \frac{1}{n}, \frac{1}{n} \right)$.
Solution: The given function is
\[
f(x,y)=
\begin{cases}
1, &\text{ if } xy = 0 ;\\
0, &\text{ if } \text{ otherwise}.
\end{cases}
\]
We note that
\begin{align*}
\frac{\partial f}{\partial x} (0,0) & = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} \\[1ex]
& = \lim_{h \to 0} \frac{1-1}{h} = 0.
\end{align*}
Similarly,
\begin{align*}
\frac{\partial f}{\partial y} (0,0) & = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} \\[1ex]
& = \lim_{k \to 0} \frac{1-1}{k} = 0.
\end{align*}
Thus, both partial derivatives of $f$ exist at $(0,0)$.
Now we will discuss the continuity of $f$ at $(0,0)$. Take a sequence $(x_n, y_n) = \left( \frac{1}{n}, \frac{1}{n} \right) $. Note that
\[
\lim_{n \to \infty} \left( \frac{1}{n}, \frac{1}{n} \right) = (0,0).
\]
Now consider
\begin{align*}
\lim_{n \to \infty} f(x_n, y_n) & = \lim_{n \to \infty} f\left( \frac{1}{n}, \frac{1}{n} \right) \\
& = \lim_{n \to \infty} 0 = 0.
\end{align*}
But, $f(0,0) = 1$ and hence, the function is not continuous at $(0,0)$.
The existence of partial derivate at a point $\mathbf{p} $ does not necessarily imply the continuity of $f$ at $\mathbf{p} $.