<s>[INST] Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$. I am new to multivariable limits and I am having a hard time arguing this limit:

Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$.

I can work out the overarching context of the proof as follows: Let $\epsilon>0$ be given. We need a $\delta$ so that whenever $\|(x,y)-(2,1)\|<\delta$ we have $\|f(x,y)-f(2,1)\|<\varepsilon.$ The difficulty I always have with these problems is that I know I need to rewrite $\|\sqrt{x+y-2}-1\|$ in terms of the given $\delta$ estimate but for some reason when it comes to these limit problems algebra turns into the most difficult thing in the world for me.
 [/INST] Let $\varepsilon > 0$ and set $\delta = \min\left\{\frac{\varepsilon^2}{6}, \frac12\right\}$. 
Let $\|(x,y) - (2,1)\| < \delta$. This implies $|x-2| < \delta$ and $|y-1| < \delta$. Also
$$x+y-2 = 1 + (x-2) + (y-1) \ge 1-|x-2|-|y-1| > 1-2\delta \ge 0$$
so $\sqrt{x+y-2}$ is well-defined.
We have
$$|x+y-2| \le |x-2| + |y-1| + 1 < 2\delta + 1$$
$$|x+y-2| \ge x+y-2> 1-2\delta$$
so
\begin{align}
\left(\sqrt{x+y-2}-1\right)^2 &= (x+y-2) - 2\sqrt{x+y-2}+1 \\
&< (2\delta+1) - 2\sqrt{1-2\delta} + 1 \\
&\le (2\delta+1) - 2(1-2\delta) + 1 \\
&\le 6\delta \\
&< \varepsilon^2
\end{align}
and hence $$\left|\sqrt{x+y-2}-1\right| < \varepsilon$$
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