question about the product of continuous functions If $f(x, y)=h(x)g(y)$ where $f:\mathbb{R}^2\rightarrow \mathbb{R}, h:\mathbb{R}\rightarrow \mathbb{R}, g:\mathbb{R}\rightarrow \mathbb{R}.$  $g, h$ are continuous functions at $y_0$  and $x_0$ respectively, then can we say that $f(x, y)$ is a continuous function at $(x_0, y_0)?$ i.e $\lim \limits_{(x, y)\rightarrow (x_0, y_0)}f(x, y)=f(x_0, y_0)?$
 A: Yes. 
Let $\epsilon>0$. Since $h$ is continuous at $x_0$, there is a $\delta_1>0$ such that $|h(x)-h(x_0)|<\epsilon$ for each $x$ such that $|x-x_0|<\delta_1$.
Notice that for each such $x$ we have $|h(x)|=|h(x)-h(x_0)+h(x_0)|\leq|h(x)-h(x_0)|+|h(x_0)|\leq\epsilon+|h(x_0)|$.
Since $g$ is continuous, there is a $\delta_2>0$ such that $|y-y_0|<\delta_2$ implies $|g(y)-g(y_0)|<\epsilon$.
Take $\delta=\min\{\delta_1,\delta_2\}$. 
Then, if $(x,y)\in\mathbb{R}^2$ such that $\|(x,y)-(x_0,y_0)\|<\delta$, we have
\begin{eqnarray}
|x-x_0| & =& \sqrt{(x-x_0)^2}\leq\sqrt{(x-x_0)^2+(y-y_0)^2}\\
& =& \|(x-x_0,y-y_0)\|=\|(x,y)-(x_0,y_0)\|<\delta\leq\delta_1,
\end{eqnarray} 
so $|h(x)-h(x_0)|<\epsilon$. Similarly, we find $|y-y_0|<\delta_2$, so $|g(y)-g(y_0)|<\epsilon.$
Then
\begin{align*}
|f(x,y)-f(x_0,y_0)| & =|h(x)g(y)-h(x_0)g(y_0)|=|h(x)g(y)-h(x)g(y_0)+h(x)g(y_0)-h(x_0)g(y_0)|\\
& \leq  |h(x)g(y)-h(x)g(y_0)|+|h(x)g(y_0)-h(x_0)g(y_0)|\\
& = |h(x)||g(y)-g(y_0)|+|h(x)-h(x_0)||g(y_0|\\
&< (\epsilon+|h(x_0)|)|g(y)-g(y_0)|+ \epsilon|g(y_0)|\\
&< (\epsilon+|h(x_0)|)\epsilon+\epsilon|g(y_0)|\\
& = \epsilon\big((\epsilon+|h(x_0)|+|g(y_0)|\big)
\end{align*}
Since $h(x_0)$ and $g(y_0)$ are constants, it follows that $f(x,y)\to f(x_0,y_0)$ if $(x,y)\to(x_0,y_0)$.
A: To prove it, use the fact that when $\{x_n\}$ and $\{y_n\}$ are sequences and $x_n \to x_0$, $y_n \to y_0$, then by continuity of $h$ and $g$, $h(x_n)\to h(x_0)$ and $g(y_n) \to g(y_0)$. This shows that $$f(x_n,y_n) = h(x_n)g(x_n) \to h(x_0)g(y_0)=f(x_0,y_0).$$ 
