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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.

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  • $\begingroup$ 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 $f(a_n)$. $\endgroup$ – Ivon Dec 18 '19 at 16:12
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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 $\lim_{n\to\infty}f(a_n)=L$.

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