Let $f:D\to\mathbb R$ and let $c$ be an accumulation point of $D$. Then
$(i)\lim_{x\to c}f(x)=L$ iff $(ii)$ for every sequence $(s_n)$ in $D$ that converges to $c$ with $s_n\neq c$ the sequence $(f(s_n))$ converges to $L$

I'm okay with one direction.

To prove the other direction (taking the contrapositive statement):

Suppose $L$ is not a limit of $f$ at $c$. Find a sequence $s_n$ in $D$ such that $s_n$ converges to $c$ but $(f(s_n))$ does not converge to $L$

Since $L$ is not a limit of $f$ at $c$, $\exists\epsilon>0$ such that $\forall\delta>0$ $\exists x\in D$ such that $0<|x-c|<\delta$ implies $|f(x)-L|\ge\epsilon$.

Now the book I'm reading, Steven Lay's "Analysis with an introduction to proof" goes on as follows:

" In particular, for each $n\in\mathbb N$, there exists $s_n\in D$ with
$0<|s_n-c|<1/n$ such that $|f(s_n)-L|\ge\epsilon$"

Thus exhibiting $(s_n)$ as the required sequence.

I'm not sure why is it required that $\delta$ must be related to $1/n$

. . .

ok, I want to show that there exists a sequence $s_n$ that converges to $c$ such that $(f(s_n))$ does not converge to $L$

Let $s_n$ coverge to $c$. Then $\forall \delta>0 \exists N\in \mathbb N$ such that $n\ge N \to |s_n-c|<\delta$

Now I want to make this statement into:

$\forall \delta>0 \exists s_n \in D$ such that $|s_n-c|<\delta$

please detail how that happens.

  • $\begingroup$ so what? I mean what does that do? why do I have to refer to a $\delta_n$ when I can just fix any delta and claim that the terms of the sequence $s_n$ are less then $\delta$ far away from c. I feel like I'm missing something Really obvious here.. $\endgroup$
    – ak87
    Feb 21, 2017 at 0:11
  • $\begingroup$ The answer to your question is in effect my question. Can you please just explain to me in detail why I will need to relate delta to 1/n? I'm ok assuming that the limit of the function exists and proving that for every sequence converging in the domain of the function, the function of the sequence also converges. $\endgroup$
    – ak87
    Feb 21, 2017 at 0:22
  • $\begingroup$ What direction are you okay? Is it $(i)\implies(ii)$? $\endgroup$ Feb 21, 2017 at 0:24
  • $\begingroup$ Okay, you are trying to use contrapositive to prove $(ii)\implies(i)$. So we need to show that $\sim (i)\implies \sim (ii)$. Yes, you are correct by assuming $\sim(i)$. Do you know how to find an equivalent statement for $\sim(ii)$? $\endgroup$ Feb 21, 2017 at 0:37
  • $\begingroup$ yes. there exists an epsilon such that for all delta 0<|x-c|< $\delta$ implies $|f(x)-L| \ge \epsilon$ $\endgroup$
    – ak87
    Feb 21, 2017 at 0:41

2 Answers 2


The reason for the $1/n$ is that he is trying to produce a sequence that, by construction, converges to $c$. It would work just as well if you replaced $1/n$ with any other sequence of positive numbers that converge to $0$. Say $1/log(n)$ or $0.5^n$.

As for the argument that you gave, it is a fine proof of a different statement. Namely that if $L$ is the limit, then all sequences must converge to $L$. But it says nothing about what happens if $L$ is not the limit. (He's trying to prove that if $L$ is not the limit, then some sequence $s_n$ converges to $c$ but $f(s_n)$ does not converge to $L$.).


This is to see if I'm getting this straight, so please critique my answer.

To show $\lnot(ii)$, I want to find a sequence $s_n$ such that $s_n$ converges to c but $f(s_n)$ does not converge to L, given $\lnot(i)$.

$\lnot(i)$ states that $\exists\epsilon>0$ such that $\forall\delta>0$ $\exists x\in D$ such that $0<|x-c|<\delta$ implies $|f(x)-L|\ge \epsilon$

Let $s_n$ be a convergent sequence such that $\forall n \in \mathbb N$ $[s_n\in D]$. Then $\forall\delta>0$ $\exists N\in \mathbb N$ such that $n \ge N$ implies $0<|s_n-c|<\delta$ (since $\forall n \in \mathbb N$ $s_n \neq c$).

Since the existence on $x \in D$ in $\lnot (i)$ is dependent on $\delta$ I now must construct the sequence so that $\lnot (i)$ holds for all $n$, rather than just $n \ge N$, which requires $\delta$ to be related to the index of the sequence.

Thus, by $\lnot (i)$, for $\delta_n=1/n$,

$\exists s_1$ such that $0<|s_1-c|<\delta_1=1$

$\exists s_2$ such that $0<|s_2-c|<\delta_2=1/2$

$\exists s_3$ such that $0<|s_3-c|<1/3$


$\exists s_n$ such that $0<|s_n-c|<1/n$

Since $\forall n \in \mathbb N$ $s_n \in D$, $\lnot (i)$ applies and $0<|s_n-c|<1/n=\delta_n$ implies $|f(s_n)-L| \ge \epsilon$


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