How to write $\varepsilon$–$\delta$ definition of $\lim\limits_{n\to\infty}f(a_n)=f(c)$? In proving the equivalence of sequential and epsilon–delta definition of continuity, the following limit
$$\lim\limits_{n\to\infty}f(a_n)=f(c)$$
comes up. Here $f:\mathbf{A}\to \mathbf{B}$ where $\mathbf{A} \subseteq \Bbb{R} \land\mathbf{B}\subseteq\Bbb{R}$, $(a_n)$ is a sequence whose elements are from $\mathbf{A}$, $c\in\mathbf{A}$.
I am aware of the meaning of 


*

*$\lim\limits_{n\to\infty}a_n=c$, 

*$\lim\limits_{x\to\infty}f(x)=K\in \mathbf{B}$, and 

*$\lim\limits_{x\to c}f(x)=f(c)$. 


Marrying these three is confusing. My best guess would be that $\lim\limits_{n\to\infty}f(a_n)=f(c)$ if and only if$$\forall\varepsilon>0\ \exists N(\varepsilon)\in\Bbb{N}^+\land\exists\delta(N)>0:\\(\forall a_n\in\mathbf{A})\left(0<|a_n-c|<\delta\right)\implies|f(a_n)-f(c)|<\varepsilon\ \forall n>N$$
which is pretty convuluted, and therefore unlikely to be true. Even if correct, it hardly paints an intuitive picture.
Is there a more pleasing definition of this limit?
 A: If you begin by assuming that $f$ is continuous at $c$ then, for any sequence $a_n \to c$ we must show that $f(a_n)\to f(c)$ i.e to show that $\lim_{{n\to}\infty} f(a_n)=f(c)$
Consider the following arguments,
For a given $\epsilon >0$ we can find  a $\delta(\epsilon) >0$ such that whenever the terms of $(a_n)$ lie in the $\delta$-neighborhood of $c$ then their respective images $f(a_n)$ will lie in the $\epsilon$-neighborhood of $f(c)$. The existence of such a $\delta$ comes from the assumption that $f$ is continuous.
But this alone does not guarantee that $\lim_{{n\to}\infty} f(a_n)=f(c)$ (Why? Think about oscillating sequences)
Thus, it is necessary to choose $N(\delta)\in\Bbb N^+$ appropriately (this is precisely where we use the fact that $a_n\to c$). This would then guarantee that the terms of the sequence $(f(a_n))$ "eventually" belong to the $\epsilon $ neighborhood of $f(c)$  
We can finally conclude, $\lim_{{n\to}\infty} f(a_n)=f(c)$
A: Since $(f(a_n))$ is just a sequence, we only need to apply the usual definition of convergence of a sequence to a limit $f(c)$.
For all $\varepsilon>0$, there exists an $N$ such that if $n>N$, then $$|f(a_n)-f(c)|<\varepsilon.$$ 
(Or did I miss something?)
