Showing a function in two variables is continuous at every point I am new to functions in multiple variables, and I'm trying to get a better understanding of the continuity of such functions. Let's take the simplest of cases: $f(x,y)=x+y$. How can I show that this function is continuous? 
 A: One way is to resort to the definition as mentioned in the previous answers. So some easy functions can always be shown to be continuous or discontinuous by making use of the definition. But for a complicated multivariable function things start getting nasty!
At such times the following might be helpful.
You can prove these results for multivariable functions,
(1) The sum, difference and product of two continuous functions is always continuous.
(2) The quotient of two continuous functions is continuous as long as the denominator is not $0$.
(3) Polynomial functions are continuous.
(4) Rational functions are continuous in their domains.
(5) If $f(x,y)$ is continuous and $g(x)$ is defined and continuous on the domain of $f$ then $g(f(x,y))$ is also continuous.
(6) Every projection map is continuous.
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Now lets make use of the above results to test the continuity of $f(x,y)=x+y$.
Here, $f$ is a polynomial function $\implies$ $f$ is continuous
Now consider the following function, $g(x,y)=\tan ^{-1}\left(\frac{xy^2}{x+y}\right)$.
Here we have a composition of two functions, $\tan ^{-1}$ and $\left(\frac{xy^2}{x+y}\right)$. 
We know that $\tan^{-1}$ is continuous on $\Bbb R$, whereas $\left(\frac{xy^2}{x+y}\right)$ is a quotient of two continuous functions and hence it is continuous as long as the denominator is not $0$ i.e. as long as $x\neq -y$
Therefore their composition, $g(x,y)=\tan ^{-1}\left(\frac{xy^2}{x+y}\right)$ is continuous on $\{(x,y)\in \Bbb R^2|x\neq -y\}$
Most of the scary looking functions that you might usually come across in your course can be dealt in a similar fashion.
A: We will show that for any point of the plane - $\mathbb{R}^2$ - and for every $\epsilon>0$, we can find a $\delta=\delta(\epsilon,x_0,y_0)>0$ such that:
$$||(x,y)-x_0,y_0)||<\delta\Rightarrow|f(x,y)-f(x_0,y_0)|<\epsilon$$
So, let $\epsilon>0$. Note that:
$$||(x,y)-(x_0,y_0)||=\sqrt{(x-x_0)^2+(y-y_0)^2}$$
Also note that:
$$\begin{align*}|f(x,y)-f(x_0,y_0)||=&|x+y-(x_0+y_0)|=|(x-x_0)+(y-y_0)|\leq\\
\leq&|x-x_0|+|y-y_0|=\sqrt{\left(|x-x_0|+|y-y_0|\right)^2}=\\
=&\sqrt{|x-x_0|^2+|y-y_0|^2+2|x-x_0||y-y_0|}\leq\\
\leq&\sqrt{(x-x_0)^2+(y-y_0)^2}+\sqrt{2|x-x_0||y-y_0|}\leq\\
\leq&\sqrt{(x-x_0)^2+(y-y_0)^2}+\sqrt{(x-x_0)^2+(y-y_0)^2}=\\
=&2\sqrt{(x-x_0)^2+(y-y_0)^2}
\end{align*}$$
since $\sqrt{x}$ is subadditive and $2|ab|\leq a^2+b^2$ -proof at the end of the answer.
Then, choosing $\delta=\frac{\epsilon}{2}$, gives:
$$|f(x,y)-f(x_0,y_0)|\leq2\sqrt{(x-x_0)^2+(y-y_0)^2}=2||(x,y)-(x_0,y_0)||<2\delta=\epsilon$$


*

*Subadditivenes of $\sqrt{x}$. Let $x,y\geq0$. We will prove that
$$\sqrt{x+y}\leq\sqrt{x}+\sqrt{y}$$
We know that:
$$\begin{align*}&\left(\sqrt{x}+\sqrt{y}\right)^2=x+y+2\sqrt{x}\sqrt{y}\geq x+y\Rightarrow\\\
\Rightarrow&\sqrt{x}+\sqrt{y}\geq\sqrt{x+y}
\end{align*}$$

*We will prove that $$2|ab|\leq a^2+b^2$$
Firstly:
$$(a+b)^2\geq0\Rightarrow a^2+b^2\geq-2ab\Rightarrow2ab\geq-(a^2+b^2)$$
Then:
$$(a-b)^2\geq0\Rightarrow a^2+b^2\geq2ab\Rightarrow2ab\leq(a^2+b^2)$$
So:
$$-(a^2+b^2)\leq2ab\leq a^2+b^2\Rightarrow2|ab|\leq a^2+b^2$$


Note also that, since we have found a $\delta>0$ which is independent of $x_0,y_0$ we have, additionally shown that $f$ is uniformly continuous over $\mathbb{R}$.
A: Hint:
It suffices to generalize to several variables in the $\epsilon,\delta$ definition. Instead of considering a 1D interval $|x-x_0|<\delta$, you use a 2D neighborhood $\|(x,y)-(x_0,y_0)\|<\delta$ where the double bars denote some norm (such as Euclidean or Manhattan).
