how to prove that the following limit doesn't exist when puting L=3 using epsilon-delta definition?

Question

I have $$\lim_{x\to 0}\sqrt{4-x}=2$$ and what I did is: let $$ |\sqrt{4-x}-2|<\epsilon $$ then $$-(\epsilon+2)^2+4<x<-(-\epsilon+2)^2+4$$ so I took $\delta=min{[-(\epsilon+2)^2+4,-(-\epsilon+2)^2+4)}]$ therefore proving the limit. But I don't understand when this would be wrong. For example if instead of using the limit as 2 I used limit as 3 then the proof would be almost the same but wrong. So my question is, in which part would be wrong and why? That's what I don't understand

Answer 1

Hint: Recall the definition of a limit $L$ of a function $f$ at $x=a.$ We say $$\lim_{x \rightarrow a} f(x)=L$$ if for all $\epsilon>0,$ there exists a $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-L|<\epsilon.$
You want to disprove $$\lim_{x \rightarrow 2} \sqrt{4-x}=3.$$ What you need to do is pick a number $\epsilon>0$ such that there is no $\delta>0$ with the property that $0<|x-2|<\delta$ implies $|\sqrt{4-x}-3|<\epsilon.$

Alternatively, use the uniqueness property of limits: If $f$ has limit $L$ at $x=a,$ then $L$ is unique. That is, if $L_1,L_2$ are limits of $f$ at $x=a,$ then $L_1=L_2.$ Try proving this fact.

Then, based on this fact, can both $2$ and $3$ be limits of $\sqrt{4-x}$ at $x=2$ ?