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If it is, would it be the same reasoning that $\sqrt{x}$ is continuous at $(0,0)$?

Vasting
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2 Answers2

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The map $\Bbb R_{\geq 0} \ni t \mapsto \sqrt{t} \in \Bbb R$ is continuous, and so is the map $$\{ (x,y) \in \Bbb R^2 \mid x \geq y\} \ni (x,y) \mapsto x-y \in \Bbb R.$$Your map $f$ is the composite of these two continuous maps, hence also continuous.

Ivo Terek
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  • Yes, but you've specified a domain that the OP did not. – Ted Shifrin Aug 27 '17 at 23:20
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    True, but I'm thinking that this is probably an exercise in an early multivariable calculus course where people usually get sloppy and assume that the domain is the largest possible (not complex) :P – Ivo Terek Aug 27 '17 at 23:22
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Let $D:=\{(x,y):x \ge y\}$ and $\epsilon >0$. Then put $ \delta=\frac{\epsilon^2}{2}$. For $(x,y) \in D$ with $||(x,y)-(0,0)||_2 < \delta$ we have:

$|f(x,y)-f(0,0)|^2=x-y=|x-y| \le |x|+|y| \le 2 ||(x,y)|| <2 \delta = \epsilon^2$,

hence

$|f(x,y)-f(0,0)| < \epsilon$.

Fred
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