For a function of two variables to be continous, a sufficient condition is said to be that one of the partial derivatives exists and is bounded near the required point and the other partial derivative just exists at that point
I cant understand how the bounded partial derivative helps in the proof.Please explain this to me Thanks in advance
Let $p=(x_0,y_0)$ be the required point. For $(x,y)$ near $p$ with $y \ne y_0$ we have
$f(x,y)-f(p)=f(x,y)-f(x_0,y)+f(x_0,y)-f(p)$.
By the mean value theorem there is $t_x$ between $x $ and $x_0$ such that
$f(x,y)-f(x_0,y)=f_x(t_x,y)(x-x_0)$, hence
$f(x,y)-f(p)=f_x(t_x,y)(x-x_0)+\frac{f(x_0,y)-f(p)}{y-y_0}(y-y_0) $.
Since $f_x$ is bounded near $p$ we have $f_x(t_x,y)(x-x_0) \to 0$ as $(x,y) \to p$.
Furthermore $\frac{f(x_0,y)-f(p)}{y-y_0}(y-y_0) \to f_y(p) \cdot 0$ as $(x,y) \to p$.
Conclusion: $f(x,y)-f(p) \to 0$ as $(x,y) \to p$.