I want to prove the definition of limit of a function from definition (1) to definition (2).

(1) A function $f(x)$ is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for every sequence $\{x_n\}$ with limit $x^*$, the limit $\lim_{n->\infty}f(x_n)$ exists and is equal to $f^*$. (2) A function is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for each $\epsilon >0$, there exists a $\delta_\epsilon >0$ such that $|f(x)-f^*| <\epsilon \forall |x^*-x|<\delta_\epsilon$

I am confused about the steps. Suppose I am given an $\epsilon > 0$ and $x^*$. I have to show the existence of a $\delta_\epsilon >0$ such that for all $x$ satisfying $|x^*-x|<\delta_\epsilon$ I can construct a sequence $x_n$ which converges to $x^*$ and $x_n=x$ for some $n \geq N$. Then I can say that $f(x_n)-f^*=f(x)-f^* < \epsilon$. I think I may have made some mistake about the steps.

My main confusion is the following : Suppose I choose $\delta_\epsilon=\epsilon$. Then if a sequence $\{x_n\}$ converges to $x^*$ then $x_n=x$ (where $x$ satisfies $|x-x^*|<\epsilon$) for some $n\geq N$. Now, $\lim_{n->\infty}f(x_n)$ exists implies there exists an $M$ such that $\forall n\geq M$, $f(x_n)-f^* <\epsilon$. Now, I need to show that for some $n \geq M$ indeed $x_n=x$. How do I show that ? This $M$ and $N$ are different.

  • $\begingroup$ The definition of limit is your $(2)$. It is a theorem that in metric spaces $(2)$ is equivalent with $(1)$. $\endgroup$ Sep 10, 2013 at 15:57
  • $\begingroup$ @ChristianBlatter: I have added some clarification and my confusion. $\endgroup$
    – aaaaaa
    Sep 10, 2013 at 18:24

1 Answer 1


If $\xi$ is an accumulation point of ${\rm dom}(f)$ then by definition $\lim_{x\to\xi}f(x)=\eta$ if, given any tolerance $\epsilon>0$, there is an allowance $\delta>0$ such that $0<|x-\xi|<\delta$ assures $|f(x)-\eta|<\epsilon$. (This is your definition $(2)$.)

Now it is a theorem that in metric spaces a single instance of $\lim_{x\to\xi}f(x)=\eta$ can be verified by testing a gogol of instances $\lim_{n\to\infty} f(x_n)=\eta$ for sequences $(x_n)_{n\geq1}$ converging to $\xi$. To be exact:

Theorem. One has $\lim_{x\to\xi}f(x)=\eta$ iff for all sequences $$n\mapsto x_n\in{\rm dom}(f)\setminus\{\xi\}\tag{1}$$ with $\lim_{n\to\infty}x_n=\xi$ one has $\lim_{n\to\infty} f(x_n)=\eta$.

Proof. Assume that $\lim_{x\to\xi}f(x)=\eta$ and that $(1)$ is such a sequence. Given an $\epsilon>0$ there is a $\delta>0$ such that $0<|x-\xi|<\delta$ assures $|f(x)-\eta|<\epsilon$, and as $\lim_{n\to\infty}x_n=\xi$ there is an $n_0$ such that $0<|x_n-\xi|<\delta$ for all $n>n_0$. It follows that $|f(x_n)-\eta|<\epsilon$ for $n>n_0$, and as $\epsilon>0$ was arbitrary it follows that $\lim_{n\to\infty} f(x_n)=\eta$.

Conversely: Assume that $\lim_{x\to\xi}f(x)=\eta$ does not hold. Then by definition there is an $\epsilon_0>0$ without corresponding $\delta>0$. In particular, for each $n\geq1$ there is a "bad" point $x_n\in{\rm dom}(f)$ with $0<|x_n-\xi|<{1\over n}$ and $|f(x_n)-\eta|\geq\epsilon_0$. The points $x_n$ so obtained constitute a sequence in ${\rm dom}(f)$ converging to $\xi$, for which $\lim_{n\to\infty} f(x_n)=\eta$ does not hold.$\quad\square$

  • $\begingroup$ For the converse proof I am interested in a straight forward proof rather than by contradiction. So, I have to show an existence of $\delta_\epsilon$ $\endgroup$
    – aaaaaa
    Sep 11, 2013 at 5:58
  • $\begingroup$ @Prasenjit: It would be nice to have such a proof. But I don't think I have ever seen one. $\endgroup$ Sep 11, 2013 at 9:42
  • $\begingroup$ @ChristianBlatter, the converse part of this proof relies on the axiom of countable choice. $\endgroup$ Jul 3, 2017 at 12:22

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