Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (0,0)}(\sqrt{-x^2-y^2})=0$

Question

Use epsilon-delta definition of limits to show that

$$\lim_{(x,y) \to (0,0)}(\sqrt{-x^2-y^2})=0$$

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{x^{2}+y^{2}}<\delta \implies \left|\sqrt{-(x^{2}+y^{2})}\right|<\epsilon)))$$

If $0<\sqrt{x^{2}+y^{2}}<\delta$ then $\sqrt{-(x^{2}+y^{2})}<0<\delta$

So I think if $\delta \le \epsilon$ then the claim does hold,is that true?

Answer 1

You have to be careful with the definition of limits in this case. The function $$ f(x,y)=\sqrt{-x^2-y^2} $$ is only defined on the set $E=\{(0,0)\}$.

So the notion of limits must be restricted to the (trivial) metric subspace $E$ of $\mathbb{R}^2$.

In the $\epsilon$-$\delta$ definition of your limit, you should have

for every $\epsilon$, there exists $\delta>0$ such that the condition $0<\sqrt{x^2+y^2}<\delta$ and $(x,y)\in E$ implies that $|f(x,y)|<\epsilon$ .

But this is vacuously true since $0<\sqrt{x^2+y^2}<\delta$ is never satisfied.

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Added notes.

In general, let $X$ and $Y$ be two metric spaces and $E\subset X$, $f:E\to Y$. To avoid the vacuously true case, one only defines the notion of limits $\lim_{x\to p}f(x)$ when $p$ is a limit point $E$, meaning that every neighborhood of $p$ contains a point $q\ne p$ such that $q\in E$. (See for instance Rudin's Principles of Mathematical Analysis.) Your example of $E$ has no limit points.