Answer to Question #104618 in Calculus for Gayatri Yadav

Question

Even if both the partial derivatives of f : R^2 → R
exist at a point (a,b) , the function f may still not be continuous at (a,b).

Answer 1

Let we have function $f: E\longrightarrow R, E\sub R^n, E \text{ is an open set}.$

Theorem. If $f$ is differentiable at a point $x_0$ f is continuous at a point $x_0$.

In case $n\geq 2$ if partial derivatives of f exist f may be not differentiable. For example, consider the function $f(x,y)=1 \text{ if } xy=0 \text{ and } f(x,y)=0 \text{ otherwise}.$ Then we have $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$ but function f is not continuous at (0,0):$lim_{(x,y)\longrightarrow (0,0)}=0\neq f(0,0)=1.$ So function f is not differentiable at (0,0).