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?
