Why if $\lim_{x \to a}f(x)=b$ then $\lim_{n \to \infty}f(x_n)=b$? Why is that if $$\lim_{x \to a}f(x)=b \Rightarrow \lim_{n \to \infty}f(x_n)=b?$$
This is often used to prove that the same properties that apply to sequences, also apply to functions, but I don't know from where this implication is coming from.
It is given that $\lim_{n \to \infty}x_n=a$ and that is derived from the fact that $a \in D'f$. I am not sure about this part either, t.i., why the $a$ being a limit point implies that there is a sequence within the domain that converges to $a$.
 A: You should note that the proper result in question is like this:

Theorem: Let $\lim_{x\to a} f(x) =b$ and let $\{x_n\} $ be any sequence such that $x_n\neq a$ after a certain value of $n$ and $\lim_{n\to\infty} x_n=a$. Then $\lim_{n\to\infty} f(x_n) =b$.

The part $x_n\neq a$ is important and the result fails if this hypothesis is not met. One can prove the theorem using the definition of limit of sequences and functions.
If you are familiar with these definitions then you should try to prove the result. Otherwise it is best to understand it informally. If $x$ is near (but not equal to) $a$ then $f(x) $ is near $b$. And if $n$ grows without bound then $x_n$ is near $a$ (and by hypothesis not equal to $a$) and hence $f(x_n) $ is near $b$.

The second part of your question is related to the definition of a limit point and if you are familiar with it then you can prove the following result

Lemma: If $a$ is a limit point of a non-empty set $A$ then there is a sequence $\{x_n\} $ such that $x_n\in A, x_n\neq a$ for all $n$ and $\lim_{n\to\infty} x_n=a$.

The idea is to consider interval $(a-1/n,a+1/n)$ which necessarily contains some $x_n\in A$ with $x_n\neq a$. 
A: If $(x_{n})_{n\in \mathbb{N}}$ is a sequence such that $\lim_{n \to \infty} x_{n} = a$
Then if $\lim_{x \to a} f(x) = b$ we have that for all $\varepsilon >0$ there exist $\delta >0$ such that if $|x-a|<\delta \implies |f(x)-b| < \varepsilon$
So, if we take a fixed $\varepsilon>0$, there exist a $\delta>0$ then by definition of limit, there exist $N\in \mathbb{N}$ such that if $n≥N$ then $|x_{n}-a|<\delta$ and that implies $|f(x_{n}) - b |<\varepsilon$
In other words, $\lim_{n\to \infty} f(x_{n}) = b$
A: What the limit says is that however you approach the point $a$, the function $f(x)$ approaches $b$. Notice however you approach. That means in particular if you take any sequence $x_n$ which approaches $a$, then $f(x_n)$ must also approach $b$.
Formally if $f(x)\to b$ as $x\to a$, then by definition for all $\varepsilon>0$ there is a $\delta > 0$ so that $|x-a|<\delta$ implies $|f(x)-b|<\varepsilon$. Now suppose $x_n$ is a sequence which converges to $a$. Then for any $\delta>0$ there is an $N$ so that $n\geq N$ implies $|x_n-a|<\delta$. But since for these $n\geq N$ we have $|x_n-a|<\delta$ we must also have $|f(x_n)-b|<\varepsilon$. This 'proves' $f(x_n)\to b$ as $n\to\infty$.
It is however very important that $x_n\to a$, you cannot take any sequence (obviously).
