Does $|x-1|<\delta\implies |f(x)-4|<\epsilon$ and so $\lim_{x\to 1} f(x)$ exists?

Question

Suppose that for any $\epsilon>0$ the solution to $|f(x)-4|<\epsilon$ is $$x\in\left(1-\frac{\epsilon}{2},1+\frac{\epsilon}{2}\right).$$ Can we conclude from this that $\displaystyle\lim_{x\to 1}f(x)$ exists? It is clear that we need to use the $\epsilon$-$\delta$ definition of the limit, where in this case the claim is that $\delta=\dfrac{\epsilon}{2}$. However, I am having trouble understanding that if $$|x-1|<\delta\implies |f(x)-4|<\epsilon$$ the limit does/does not exit. In general, my proofs involve trying to write $|f(x)-4|$ in the form of $|x-1|$ in order to find a satisfactory $\delta$. I am not sure where to start for this question.

Answer 1

If the assertion is:

For every $\epsilon>0$ we know that $$x\in\left(1-\frac\epsilon2,1+\frac\epsilon2\right) \implies |f(x)-4|<\epsilon$$

then yes, it is the case that $\lim_{x\to 1}f(x)$ exists and equals $4$. The $\epsilon-\delta$ proof goes as follows. Let $\epsilon>0$. Choose $\delta:=\frac\epsilon2$. If $|x-1|<\delta$ then $x\in(1-\frac\epsilon2,1+\frac\epsilon2)$ so by the assertion, $|f(x)-4|<\epsilon$.

OTOH, if the assertion is:

For every $\epsilon>0$ there exists $x\in\left(1-\frac\epsilon2,1+\frac\epsilon2\right)$ such that $|f(x)-4|<\epsilon$

then no, this is not enough to show that $\lim_{x\to 1}f(x)$ exists and equals $4$. Can you see a counterexample?