Prove $f(x,y) =\sqrt{16-x^2-y^2}$ is continuous using $\epsilon$, $\delta$? I want to prove that for every $\epsilon > 0$ there exists a $\delta > 0$ such that
$$(x-x_0)^2 + (y-y_0)^2 < \delta^2 \Rightarrow  \left|\sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right| < \epsilon.$$
I note that $|x-x_0| \leq \delta$ and $|y-y_0| \leq \delta$.
What I've tried:
$$\left| \sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right| = \frac{|x^2-x_0^2 + y^2-y_0^2|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} }$$
$$= \frac{|(x-x_0)(x+x_0) + (y-y_0)(y+y_0)|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} } \leq 
\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} }$$
$$\leq 
\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x_0^2-y_0^2} }$$
But then I do not know how to continue.
 A: HINT
You are on the right track! Notice that
\begin{align*}
\begin{cases}
|x + x_{0}| = |(x - x_{0}) + 2x_{0}| \leq |x - x_{0}| + 2|x_{0}|\\\\
|y + y_{0}| = |(y - y_{0}) + 2y_{0}| \leq |y - y_{0}| + 2|y_{0}|
\end{cases}
\end{align*}
Consequently, if we set $k^{-1} = \sqrt{16 - x^{2}_{0} - y^{2}_{0}}$, then we have that
\begin{align*}
k(|x - x_{0}||x + x_{0}| + |y - y_{0}||y + y_{0}|) \leq 2k\delta^{2} + 2k\delta(|x_{0}| + |y_{0}|) := \varepsilon
\end{align*}
Can you take it from here?
A: First note that if $|x-x_0|+|y-y_0|<1$ then each of $|x+x_0|,\ |y+y_0|$ is bounded by some $M>0\ $ (Proof: triangle inequality). So assume $\delta<1.$  There are two cases:
$1).\ $ If $x_0^2+y_0^2\neq16,\ $ set $N=\sqrt{16-x_0^2-y_0^2}.$ Then
$\displaystyle\left|\sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right|\le \frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x_0^2-y_0^2}}=\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{N}<\frac{M}{N}\cdot (|x-x_0|+|y-y_0|)$
and we may take $\delta=\min\{1,\epsilon N/M\}.$
$2).\ $ If $x_0^2+y_0^2=16$ then
$\displaystyle|f(x,y)-f(x_0,y_0)|^2=16-x^2-y^2=x_0^2-x^2+y_0^2-y^2<M(|x-x_0|+|y-y_0|)$ so we may take $\delta=\min\{1,\epsilon^2/M\}.$
