Theorem that links limits of successions and limits of functions I've a doubt on the theorem that links the limits of functions with the limits of successions.
$$\lim_{x\to c}f(x)=l \iff \forall \big\{x_n\big\}_{x\in \mathbb{N}} \mid  x_n\in dom(f) \wedge x\neq c \wedge \lim_{n\to \infty} x_n = c$$ holds that $\lim_{n\to \infty} f(x_n)=l$.
Must the condition $x_n\neq c$ be true $\forall x_n$ or it just means that the considered succession cannot be a constant one? (i.e. $x_n=c$ $ \forall x_n$)
Thanks for your help

Yes, the first one: "The condition must be true for every $x_n$". Of course this excludes in particular constant sequences ($x_n=c$). For example let $$f(x)=\begin{cases}0, & x=3 \\ x, & \text{ else} \end{cases}$$ Then if you take the sequence $x_n=3$ for all $n$ "converges" (trivially) to $x_0=3$, so $$\lim_{n\to \infty}f(x_n)=\lim_{n\to \infty}f(3)=0$$ But 
$$\lim_{x\to 3} f(x)=\lim_{x\to 3}x=3 \neq 0$$ So, the limit of $f$ as $x$ approaches $x=3$ exists and is equal to $3$ which is different from $f(3)=0$ but if you take the constant sequence $x_n=3$ you cannot see this.



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*The previous example is a constant sequence $x_n=c$ for every $n\in \mathbb N$. But consider also the sequence $$x_n=\begin{cases}3, & n=\text{ even}\\3+\frac1n, & n=\text{ odd}\end{cases}$$ This is a non-constant sequence with $x_n\to 3$ as $n\to \infty$ but exhibits the same problem as above. 

