# Limits of special Sequences

Suppose $$f(x)$$ be a function such that $$\lim_{x \to c}f(x)=L$$ and let $$a_n$$ be a sequence such that the limit of this sequence is $$c$$ then $$\lim_{a_n \to c}f(a_n)=$$ Here I want to know that for the limit of $$f$$ to be defined as $$x \to c$$ $$f$$ must be defined in some open interval containing $$c$$ but $$f(a_n)$$ is only defined for some special values $$a_n$$ then how the limit of $$f(a_n)$$ is defined.

• When you take limit of $f(x)$, it is with regard to the variable $x$, for example as $x\to c$, but when you take limit of $f(a_n)$, it is with regard to the natural number $n$, usually as $n\to\infty$. The sequence $a_n$ does not need to cover a neighbourhood of $c$. It is just approximating $c$ as $n\to \infty$ so that the limit of $f(x)$ at $c$ can be equivalently defined as the limit of $f(a_n)$. – Ivon Dec 18 '19 at 16:12

You have $$\tag1 \lim_{n\to\infty}f(a_n)=L.$$ From $$a_n\to c$$, $$\tag2 \forall\varepsilon>0,\ \exists N>0:\ n>N\ \implies\ |a_n-c|<\varepsilon.$$ From $$(1)$$ you have $$\tag3 \forall\varepsilon>0,\ \exists \delta>0:\ |x-c|<\delta\ \implies\ |f(x)-L|<\varepsilon.$$ Now, given $$\varepsilon>0$$, take $$\delta>0$$ as in $$(3)$$. Use this $$\delta$$ as the "$$\varepsilon$$"in $$(2)$$ to get an $$N$$. So if $$n>N$$ we have that $$|a_n-c|<\delta$$, and so $$|f(a_n)-L|<\varepsilon$$. So we have shown that $$\tag4 \forall\varepsilon>0,\ \exists N>0:\ n>N\ \implies\ |f(a_n)-L|<\varepsilon,$$ which is precisely $$\lim_{n\to\infty}f(a_n)=L$$.