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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, $|\sqrt{x}-L| <\epsilon \Rightarrow L-\epsilon<\sqrt{x}<L+\epsilon,$ 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?

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

2 Answers 2

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Hint. Part of the definition of a limit says "... for all $x$ within $\delta$ of $x_0$ ..."

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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 : $\forall r>0, \ \exists y\neq x_0 \in E, \ |x_0-y|<r$.

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 $\{x_0, |x_0 - y|<r\} \cap X = \{x_0\}$. You can't approach the point of your function.

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