# Removing that $x_0$ must be an accumulation point from the definition of limit.

The textbook "Elementary Real Analysis" by Thomson, Bruckner suggests that the definition of $$lim_{x\to x_0}f(x)$$ requires that $$x_0$$ is an accumulation value of the domain of that function. One of the exercises, which I've been struggling with for a while now, states that if you remove this requirement, it is possible to arrive at circumstances in which the limit of a function at a point can be any real number. Specifically, it says to prove that, if this requirement is removed, $$lim_{x\to -2}\sqrt{x}=L$$ is possible for any real number $$L$$. Can anyone help me prove this?

I have that since $$x\ge0,|x+2|< \delta \Rightarrow 0\le x<\delta-2,$$ so if $$\delta=(L+\epsilon)^2+2,|x+2|<\delta \Rightarrow x<(L+\epsilon)^2.$$ Then, $$|\sqrt{x}-L| <\epsilon \Rightarrow L-\epsilon<\sqrt{x} and using the definition I gave for $$\delta$$, I get $$\sqrt{x}<\sqrt{\delta}=L+\epsilon,$$ providing one half of the inequality. But what about the other?

• The “if” part of the limit becomes vacuous, so is true by default. Commented May 20 at 19:48

Hint. Part of the definition of a limit says "... for all $$x$$ within $$\delta$$ of $$x_0$$ ..."
You can show that $$[0, \infty[$$ is closed and has only accumulation points.
$$x_0 \in X$$ is an accumulation point of $$E \subset X$$ if and only if : $$\forall r>0, \ \exists y\neq x_0 \in E, \ |x_0-y|.
If $$x_0$$ is not an accumulation point of the domain of your fonction $$X$$, we say it's an isolated point, means that there exists $$r>0$$, such that $$\{x_0, |x_0 - y|. You can't approach the point of your function.