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My book mentions the following theorem;

Let $f(n)$ be a continuous function defined on $[1,\infty]$ and $a_n$ be a sequence of real numbers such that $f(n)=a_n$ whenever $n$ is a positive integer, then: $\lim_{n \to \infty} f(n)=\lim_{n \to \infty} a_n$

I am working on a problem which would be simplified if the converse of the theorem holds. I tried to prove/disprove myself but I couldn't. Does the following theorem hold? How could it be proven/disproven?

Let $f(n)$ be a continuous function defined on $[1,\infty]$ and $a_n$ be a sequence of real numbers such that $f(n)=a_n$ whenever $n$ is a positive integer, then: $\lim_{n \to \infty} a_n=\lim_{n \to \infty} f(n)$

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    $\begingroup$ Well remember x=y iff y=x $\endgroup$ – Kamster Dec 30 '14 at 17:04
  • $\begingroup$ Are you familiar with the $\epsilon-\delta$ definition of limits? $\endgroup$ – Tim Raczkowski Dec 30 '14 at 17:09
  • $\begingroup$ @TimRaczkowski Yes I am..I am able to follow the proof of the original theorem but I don't know how to approach its converse. $\endgroup$ – Nesh Dec 30 '14 at 17:09
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Your second theorem is in no sense a converse to the first. It is just a restatement with an equivalent conclusion; if $a = b$ then $b = a$. One possible converse of your first theorem is

Let $f$ be a continuous function defined on $[1, \infty)$ and let $a_n$ be a sequence of real numbers. If $\lim_{n\to\infty} f(n) = \lim_{n\to\infty} a_n$ then $f(n) = a_n$ for all integers $n$.

This not true in general. For any convergent sequence $a_n$, let $b_n = a_n + \frac{1}{n}$ and let $f$ interpolate piecewise-linearly the points $\{(n, b_n):n\in\mathbb{N}\}$. Then $f$ is continuous, but $f(n) \ne a_n$ for any $n$, because $f(n) = b_n$ and $b_n \ne a_n$. However, $$\lim_{n\to\infty} f(n) = \lim_{n\to\infty} b_n = \lim_{n\to\infty} a_n + \frac{1}{n} = \lim_{n\to\infty} a_n + \lim_{n\to\infty} \frac{1}{n} = \lim_{n \to\infty} a_n + 0 = \lim_{n\to\infty} a_n.$$

Update: Perhaps you meant to compare the limits $$\lim_{x\to\infty} f(x)~\text{ and }\lim_{n\to\infty} a_n$$ rather than $$\lim_{n\to\infty} f(n)~\text{ and }\lim_{n\to\infty} a_n.$$ Indeed, your first theorem is trivial to prove as stated, and does not require continuity in the least: if $f(n) = a_n$ then the sequences $(f(n))_{n=1}^\infty$ and $(a_n)_{n=1}^\infty$ are the same so they have the same limiting behaviour. The fact that $\lim_{x\to\infty} f(x)$ is equal to $\lim_{n\to\infty} f(n)$ follows from the nontrivial fact that if $f$ is continuous, then $f$ preserves sequences.

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  • $\begingroup$ Thanks for your valuable answer. $\endgroup$ – Nesh Dec 30 '14 at 17:21

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