Multivariable continuity in ball How do you show continuity for ball-based functions such as
$$f:B[(0,0),1)]\rightarrow\mathbb{R}, \space f(x,y) = \sqrt{(1-(x^2+y^2)}$$
 A: A direct proof: Denote our ball by $B$. We observe that $f(B)=[0,1]$. Let $0 \leq a < b \leq 1$. Define $I = (a,b)$ (open interval) and consider $A = f^{-1}\left( I \right) $. Observe that $$A = \left\{x \in \mathbb{R}^2 \: \middle| \: \sqrt{1- a} <  \left| x \right| < \sqrt{1- b} \right\}$$ which is open. Thus the preimage of an open interval (and consiquently an open set) is open and $f$ is continuous.
You could also say that $f = g \circ h$, where $g(x) = \sqrt{x}$ and $h(x) = 1- (x^2+y^2)$. Thus $f$ is a composition of functions that are continuous  in their domain and thus continuous itself.
A: If you want an $\varepsilon$-argument:
Let $\varepsilon > 0$; let $(x',y') \in \mathbb{R}^{2}$ such that $x'^{2} + y'^{2} < 1$; then
$$
|\sqrt{1-x^{2}-y^{2}} - \sqrt{1- x'^{2}-y'^{2}}| \leq \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x^{2}-y^{2}} + \sqrt{1-x'^{2}-y'^{2}}} < \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x'^{2}-y'^{2}}}.
$$
If $|x-x'|, |y-y'| < 1$, then $2x'- 1 < x+x' < 2|x'| +1$ and $2y'- 1 < y+y'< 2|y'| + 1$, so $|x+x'| < \varepsilon_{1} := \max \{|2x'-1|,  2|x'| + 1\}$ and $ |y + y'| < \varepsilon_{2} := \max \{ |2y' - 1|, 2|y'| + 1 \}$, and hence
$$
\frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x'^{2}-y'^{2}}}
<
\frac{|x-x'|\varepsilon_{1} + |y-y'|\varepsilon_{2}}{\sqrt{1-x'^{2}-y'^{2}}},
$$ 
which is $< \varepsilon$ if in addition we have $|x-x'| < \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{1}}$ and $|y-y'| < \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{2}}$; we have proved that if $x,y$ are such that $x^{2}+y^{2}<1$, $|x-x'| < \min \{ 1, \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{1}} \}$, and $|y-y'| < \min \{ 1, \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{2}} \}$, then $|\sqrt{1-x^{2}-y^{2}} - \sqrt{1-x'^{2} - y'^{2}}| < \varepsilon.$
