I would just like someone to check this proof for me. I am just learning this material so I would like to make sure I am doing the proofs properly.

Question: Show that if $\displaystyle A \subset R^n $ and $\displaystyle f: A \rightarrow R $ and if the partial derivatives $\displaystyle \partial f_{x_i} $ exist and are bounded in a neighborhood of $\displaystyle \vec{a} \in R^n $, then f is continuous at $\displaystyle \vec{a} $

Proof: We have to show that $\displaystyle |f(\vec{a} + t\vec{e_i}) - f(\vec{a})| \rightarrow 0 $ as $\displaystyle t \rightarrow 0 $ where $\displaystyle t $ is a real number and $\displaystyle \vec{e_i} $ is any basis vector for $\displaystyle R^n $.

Let $\displaystyle M $ be an upper bound for all the partial derivatives of f, in some neighborhood of $\displaystyle \vec{a} $. Observe that $\displaystyle 0 \le \lim_{t \rightarrow 0} |f(\vec{a} + t\vec{e_i}) - f(\vec{a})| = \lim_{t \rightarrow 0} |t| |\frac{f(\vec{a} + t\vec{e_i}) - f(\vec{a})}{t}| = \lim_{t \rightarrow 0} |t| \lim_{t \rightarrow 0} |\frac{|f(\vec{a} + t\vec{e_i}) - f(\vec{a})}{t}| $ $\displaystyle = \lim_{t \rightarrow 0} |t| \partial f_{x_i} \le \lim_{t \rightarrow 0} |t| M = 0 $ and so $\displaystyle |f(\vec{a} + t\vec{e_i}) - f(\vec{a})| \rightarrow 0 $ as desired. QED