continuous of a function Let $U$ be a non-empty open set in $R^2$ and let $f:U\to R$ be a function. Suppose that the first partial derivactives of $f,f_1,f_2$ are defined and bouned on all of $U$. Show that $f$ is continuous on $U$
 A: Hint
Let $q$ be a point in $U$. 
We'll prove that $f\in C^0(q)$, by showing that if there is a sequence $q_n$ in $U$ that converges to $q$ then the sequence $f(q_n)$ converges to $f(q)$. This is an equivalent formulation  to the $\mathbf{\varepsilon\!-\!\delta}$ 's definition  of ${\Bbb{R}}^2$-continuity.
Solution
Let $q_n=(x_n,y_n)$ a sequence in $U$ such that  to $q_n\to q$.
Let $\varepsilon>0$ be.
From the mean valued theorem, for $f$ we can assemble:
$$f(x,b)-f(a,b)=\frac{\partial f}{\partial x}|_{(\xi_1,b)}(x-a),$$
and 
$$f(a,y)-f(a,b)=\frac{\partial f}{\partial y}|_{(a,\xi_2)}(y-b),$$
for some $|\xi_1-a|\le|\xi-a|$ and $|\xi_2-b|\le|y-b|$. 
Then
$$f(x_n,y_n)-f(a,b)=f(x_n,y_n)-f(x_n,b)+f(x_n,b)-f(a,b)
$$
$$
=\frac{\partial f}{\partial x}|_{\xi_1,b)}(x_n-a)
+
\frac{\partial f}{\partial y}|_{x_n,\xi_2)}(y_n-b)
$$
then 
$$|f(x_n,y_n)-f(a,b)|\le M_1|x_n-a|+M_2|y_n-b|$$
We can choose large enough $K$ such that $\forall n\ge K$
we have:
$$|x_n-a|<\frac{\varepsilon}{2M_1}\quad\mbox{and}\quad |y_n-b|<\frac{\varepsilon}{2M_2},$$
for some bounds $M_1,M_2$ of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ respectively.
So
$$|f(x_n,y_n)-f(a,b)|<\varepsilon,$$
i.e. $f(q_n)$ converges $f(q)$.
