Requirement for $x$ to be different from $c$ in a theorem regarding limits Consider the following theorem on limits.
$\lim_{x\to c}f(x)=l \iff \forall \big\{x_n\big\}_{x\in \mathbb{N}} \mid  x_n\in \operatorname{dom}(f) \wedge x\neq c$  $ \forall x_n\wedge \lim_{n\to \infty} x_n = c$ 
holds that $\lim_{n\to \infty} f(x_n)=l$.
I can't see why the condition $x_n\neq c$ must be true $\forall x_n$. Usually in limits the point to which the $x$ converges is excluded because the function could even not be defined in $c$, but here, since it is required for $x_n$ also to be in $\operatorname{dom}(f)$, isn't it unnecessary?
Maybe there is another reason that I don't see
Thanks for your help
 A: The condition is necessary if limit is to mean what we want it to mean.
Let $f(x)=x^2+3$ when $x\ne 0$, and let $f(0)=88$. We want to have $\lim_{x\to 0}f(x)=3$. If we allow sequences $(x_n)$ such that $x_k=0$ for infinitely many $k$ (and $x_k\ne 0$ for infinitely many $k$), then $\lim_{n\to\infty}f(x_n)$ will not exist.  
A: In words:

The limit $\lim_{x\to c} f(x) = l$ if and only if for any sequence
  $\{x_n\}_{n \in \mathbb N}$ such that each $x_n$ is in the domain of
  $f$, and $x_n \neq c$ for any $n$, and $\lim_{n\to \infty} x_n = c$ it
  is true that $\lim_{n\to \infty} f(x_n) = l$.

Consider the following function:
$$f(x) = \begin{cases} 1 & \text{ if $x = 0$ }\\ 0 & \text{ otherwise} \end{cases}.$$
See that $$\lim_{x\to 0} ~ f(x) = 0.$$ This is why we exclude $x_n = c$. The usual definition of $\lim_{x\to c} f(x)$ is that for every $\epsilon> 0$, there exists a $\delta > 0$ such that whenever $0 < |x-c| < \delta$, we have $|f(x)-l| < \epsilon$. Here the $0<|x-c|$ part is what you are missing.
