Proof for the following limit

Question

what is the proof of the limit of $$f(x)=\begin{cases}x&x\neq1\\-2 &x=1\end{cases}$$ at $x=1$?

I have attempted to prove this with the $\delta\text{-}\epsilon$ definition but have not been successful.

$$\forall\epsilon>0\space\exists\delta>0\,\text{s.t.}\space \forall x,\\0<|x-a|<\delta\Longrightarrow|f(x)-L|<\epsilon\\|x-1|<\epsilon=\delta\\0<|x-1|<\delta\ \quad \blacksquare$$

Answer 1

$$\forall\epsilon:0<|x-1|<\delta=\epsilon\implies|f(x)-1|<\epsilon$$ is a true statement. You can conclude.

---

Also ponder that

$$\forall\epsilon:|x-1|<\delta=\epsilon\implies|f(x)-1|<\epsilon$$ is a false statement.