Answer
The existence and value of a limit of $f(x)$ as $x\to c$ never depends on what happens at $x=c$.
Work Step by Step
The existence and value of a limit of $f(x)$ as $x\to c$ never depends on what happens at $x=c$. It is all written in the precise definition of the limit.
We say that $\lim_{x\to c}f(x)=L$ if for any values $\epsilon\gt0$, there is a corresponding value of $\delta\gt0$ such that for all $x$ $$0\lt |x- c|\lt \delta\Rightarrow|f(x)-L|\lt\epsilon$$
The inequallity $0\lt |x- c|\lt \delta$ tell us that $|x-c|\ne0$, meaning that $x\ne c$. What happens at $x=c$ is never of concern to the limit of $f(x)$ as $x\to c$.
For example, considering this function $$f(x)=\frac{x^2-1}{x-1}$$
- At $x=0$: $\lim_{x\to0}f(x)=f(0)=\frac{0^2-1}{0-1}=\frac{-1}{-1}=1$.
- At $x=1$: $\lim_{x\to1}f(x)=\lim_{x-1}(x+1)=1+1=2$ but $f(1)$ is not defined.
Another example: $f(x)=2$ for $x\lt 2$ and $f(x)=5$ for $x\ge2$.
Here $f(2)$ is defined and equals $5$, but $\lim_{x\to2}f(x)$ does not exist, because $f(x)$ does not approach any single number as $x\to2$.
Yet, if we have a function like this: $f(x)=2$ for $x\ne2$ and $f(x)=5$ for $x=2$
Here both $f(2)$ and $\lim_{x\to2}f(x)$ exist, but while $f(2)=5$, $\lim_{x\to2}f(x)=2$
