I don't think the confusion here is about limits. I think it's a question about functions. $f(x)$ is not a function. $f: \mathbb R \to \mathbb R$ is a function. What it does, is assigns every real number something in $\mathbb{R}$, uniquely.
For example, your constant function is a function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x)=5$ for all $x \in \mathbb{R}$. In other words, if you give me a real number, say $5$, I will return $3$, since $f(5)=3$.
THE limit of a function does not have any coherence. There are many limits one could take. For example, some limits of the constant function: $\lim_{x \to 0}f(x)=3=\lim_{x \to 1} f(x)=3= \lim_{x \to 2} f(x)=3$ and so on. In other words, the limit here is trivial. Take $\lim_{x \to 0}f(x)$. We want to $f$ to get sufficiently close to $3$. In our case, we need not look far, as for all $x \in \mathbb R$, $f(x)=3$, and so $|f(x)-3|=0$.
As for the last part of your question, limits are indeed unique.
"proof" suppose that $\lim_{x \to y} f(x)=a$, but also that $\lim_{x \to y} f(x)=b$ for $a \neq b$. Then, on what interval can we ensure that $f(x)$ converges? The problem here is that what this limit tells us is that we can get arbitrarily close to different values, but this can't be the case! Take the distance $\frac{a-b}{2}$. Here, even as $x \to y$, $f(x)$ can't get close enough to one of them.
To perhaps aid in intuition, take $g: \mathbb R \to \mathbb R$ defined by $g(x)=\frac{1}{x}$. perhaps you've noticed the asymptote, or the proof that $\lim_{x \to \infty} \frac{1}{x}=0$. Could it be the case that $\lim_{x \to \infty} \frac{1}{x}=-1$? Or any negative number for that matter? Then there would always be the distance between that number and the $x-axis$, a distance that $\frac{1}{x}$ can never satisfy.
I'll leave the $\epsilon-\delta$ to another answer, I just wanted to maybe help untangling this shady business in analysis, without deferring to rigour.