# Precise proof of equivalence of two definitions of limit of a function

Consider two (equivalent) definitions of a (real) limit of a function $f:\mathbb{R}\rightarrow\mathbb{R}$.

The epsilon-delta one: $$\lim_{x\to x_0} f(x)=l \iff \forall \varepsilon>0\exists \delta>0\forall x\in D_f: (0<|x-x_0|<\delta \Rightarrow |f(x)-l|<\varepsilon)$$ and the sequential one: $$\lim_{x\to x_0} f(x)=l \iff \forall(x_n)\subseteq (D_f\setminus\{x_0\}):(\lim_{n\to\infty}x_n=x_0 \Rightarrow \lim_{n\to\infty}f(x_n)=l)$$ where $(x_n)\subseteq (D_f\setminus\{x_0\})$ stands for a number sequence that takes values only from the domain of $f$ without $x_0$.

I have spent hours trying to come up with or to find an axiomatic proof of their equivalence. I have seen a few different proofs, but they all seem to be imprecise, at some point just "stating the obvious".

So, is there a book, an article, a site, a lecture or something, where I can get (maybe long and boring, but still) a formal proof of this fact? Or maybe precise proofs for this fact need more advanced theory, which is not even in my books yet? Maybe I need a better understanding of what is "precise"?

• It would probably be helpful if you posted exactly what steps of these proofs you find to be imprecise. Or posted your own attempt at a proof and told us where you find it to be lacking. – kccu Oct 30 '16 at 21:59
• @kccu The main reason I did not include them is that they largely consist of non-english language. So it would take me hours to rewrite them in English. I might try rewriting my own attempts (there are less words involved) and attaching photos of them, but that is also going to take time. – zoickx Oct 30 '16 at 22:11
• I have posted my own proof as an answer. Please tell me if there is anything you find "imprecise" about it. – kccu Oct 30 '16 at 22:34

Here is my own proof. Please tell me what if any steps of this seem "imprecise" to you.

If $f$ has an "epsilon-delta limit" of $l$ at $x_0$, then $f$ has a "sequence limit" of $l$ at $x_0$: Let $(x_n)$ be a sequence in $D_f\setminus\{x_0\}$ such that $\displaystyle\lim_{n \to \infty} x_n=x_0$. We need to show that $\displaystyle\lim_{n \to \infty} f(x_n)=l$. Let $\epsilon>0$. Then, since $f$ has an "epsilon-delta limit" of $l$ at $x_0$, there exists $\delta>0$ such that for all $x \in D_f$ satisfying $0 <|x-x_0|<\delta$, we have $|f(x)-l|<\epsilon$. Since $\displaystyle\lim_{n \to \infty}x_n=x_0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|x_n-x_0|<\delta$. Then $0<|x_n-x_0|<\delta$ for all $n \geq N$, so $|f(x_n)-\ell|<\epsilon$ for all $n \geq N$. We have shown that given $\epsilon>0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|f(x_n)-l|<\epsilon$. Thus $\displaystyle\lim_{n \to \infty}f(x_n)=l$.

If $f$ has a "sequence limit" of $l$ at $x_0$, then $f$ has an "epsilon-delta limit" of $l$ at $x_0$: We will prove the contrapositive, i.e., if $f$ does not have an "epsilon-delta limit" of $l$ at $x_0$, then $f$ does not have a "sequence limit" of $l$ at $x_0$. If $f$ does not have an "epsilon-delta" limit of $l$ at $x_0$, then there exists $\epsilon>0$ such that for all $\delta>0$, there exists $x \in D_f$ satisfying $0<|x-x_0|<\delta$, and yet $|f(x)-l| \geq \epsilon$. To show that $f$ does not have a "sequence limit" of $l$ at $x_0$, we must exhibit a sequence $(x_n)\in D_f \setminus\{x_0\}$ such that $\displaystyle\lim_{n \to \infty}x_n=x_0$ and yet $\displaystyle\lim_{n \to \infty}f(x_n) \neq l$.

We construct the sequence as follows. Setting $\delta_1=1$, there exists $x_1\in D_f$ with $0<|x_1-x_0|<\delta_1=1$ such that $|f(x_1)-l| \geq \epsilon$. (Note then that $x_1 \neq x_0$, so in fact $x_1 \in D_f \setminus\{x_0\}$.) Similarly, setting $\delta_2=\frac{1}{2}$, there exists $x_2 \in D_f$ with $0<|x_2-x_0|<\delta_2=\frac{1}{2}$ such that $|f(x_2)-l| \geq \epsilon$. (Once again, $x_2 \neq x_0$, so $x_2 \in D_f \setminus\{x_0\}$.) Now suppose we have chosen $x_1,x_2,\dots,x_n \in D_f \setminus\{x_0\}$ such that $|x_i-x_0|<\frac{1}{i}$ and $|f(x_i)-l| \geq \epsilon$ for all $1 \leq i \leq n$. Setting $\delta_{n+1}=\frac{1}{n+1}$, there exists $x_{n+1} \in D_f$ with $0 \leq |x_{n+1}-x_0|<\frac{1}{n+1}$ such that $|f(x_{n+1})-l| \geq \epsilon$, so we let $x_{n+1}$ be the next term of the sequence. By the axiom of dependent choice, we can define a sequence $(x_n) \in D_f \setminus\{x_0\}$ such that $0<|x_i-x_0|<\frac{1}{i}$ but $|f(x_i)-l| \geq \epsilon$ for all $i \geq 1$.

Claim: $\displaystyle\lim_{n \to \infty} x_n=x_0$. Let $\eta>0$. Then there exists a natural number $N>\frac{1}{\eta}$. For all $n \geq N$, $|x_n-x_0|<\frac{1}{n}\leq\frac{1}{N}< \frac{1}{1/\eta}=\eta$. So $\displaystyle\lim_{n \to \infty} x_n=x_0$.

Claim: $\displaystyle\lim_{n \to \infty} f(x_n)=x_0$. Let $\epsilon$ be as above. Choose any $M \in \mathbb{N}$. Then for any $m \geq M$, $|f(x_m)-l| \geq \epsilon$. (Note: this is stronger than we need. We only need to show there exists $m \geq M$ satisfying that equation.) Hence $\displaystyle\lim_{n \to \infty} f(x_n) \neq l$.

It is good to be careful when one is first learning to write proofs, but as we can see above, sometimes including all the details makes proofs long and tedious. For instance, if I were writing this proof elsewhere, I would said that it is clear that the sequence $(x_n)$ satisfying $|x_i-x_0|<\frac{1}{i}$ for all $i$ converges to $x_0$. I would have also said it's clear that $\displaystyle\lim_{n \to \infty} f(x_n) \neq l$ is clear if each term $f(x_i)$ satisfies $|f(x_i)-l| \geq \epsilon$. And when defining that sequence $(x_n)$ inductively, I might not have written out the whole induction argument, but rather said "and so on..." These sorts of shortcuts are not necessarily imprecise. They are just an indicator that the fact that is claimed is routine to verify, and so the author has not taken the time and space to spell out all the details.

• Minor edits: "$x_1 \in D_f$" $\rightarrow$ "$x_1 \in D_f \smallsetminus \{x_0\}$ and similarly for $x_2$. Also "$|geq$" $\rightarrow$ "$\geq$". – Eric Towers Oct 30 '16 at 22:40
• @EricTowers Fixed the $\geq$. I purposely did not write $x_1 \in D_f \setminus\{x_0\}$ because negating the epsilon-delta definition of limit tells us there exists $x_1$ (in the domain of $f$) satisfying $0<|x_1-x_0|<\delta_1$ yet $|f(x_1)-l| \geq \epsilon$. One can discern from this that in fact $x_1 \neq x_0$, so $x_1 \in D_f \setminus \{x_0\}$. I suppose I took a bit of a shortcut by not writing that sentence down. – kccu Oct 30 '16 at 22:44
• My deleted answer (you beat me to it) did work through that caveat, transforming $0 \leq |x_i - x_0| < \delta$ into $0 < |x_i - x_0| < \delta$. This is the only reason its absence caught my eye in your proof. – Eric Towers Oct 30 '16 at 22:51
• The existence of the sequence $(x_n)_n$ follows from the axiom of Dependent Choice, a corollary of the axiom of Choice, not from the principle of Induction. But I give it a $+1.$ – DanielWainfleet Oct 30 '16 at 22:51
• @user254665 Could you elaborate and tell me why induction is not sufficient to prove the existence of the sequence? I don't disbelieve you, I just want to know where my reasoning fails. – kccu Oct 30 '16 at 22:59

Call the epsilon-delta def'n "$I$" and the sequential def'n "$II$".

To show $I\implies II:$ Let $(x_n)_n$ be any sequence in $D_f$ converging to $x.$ For any $e >0,$ take $d_e >0$ such that $$(\bullet ) \;\forall y\in D_f\cap (-d_e+x,d_e +x)\; ( |f(y)-l|<e).$$ Now $\lim_{n\to \infty}x_n=x$ implies that $\{n: x_n\not \in (-d_e+x,d_e+x)\}$ is finite, so $$\exists m_e\;\forall n\geq m_e\; (x_n\in D_f\cap (-d_e+x,d_e+x)\;).$$ Applying $(\bullet )$ to this we have $$\exists m_e\; \forall n\geq m_e\; (|f(x_n)-l|<e).$$ This holds for any $e>0,$ so $(f(x_n))_n$ converges to $l.$

To show $(\neg I) \implies (\neg II):$ The negation of $I$ is $$\exists e>0\;\forall d>0\;\exists y\in D_f\cap (-d+x,d+x)\;(|f(y)-l|\geq e.$$ Take such an $e.$ For each $n\in \mathbb N$ let $d_n=2^{-n}$ and take $x_n\in D_f\cap (-d_n+x,d_n+x)$ such that $|f(x_n)-l|\geq e.$ Then $(x_n)_n$ is a sequence in $D_f$ converging to $x,$ and $(f(x_n))_n$ does not converge to $l,$ so we have $(\neg II).$

• The two answers preceding mine appeared as I was typing,de-bugging, and re-writing,and re-de-bugging. And I am an extremely slow typist. – DanielWainfleet Oct 30 '16 at 22:48

We have (from the definition of limit of a sequence) that

$$\lim_{n\rightarrow\infty} x=x_0$$

means there exists for all $\delta>0$ a $N$ such that for $n> N$

$$|x_n-x_0|<\delta,$$

and

$$\lim_{n\rightarrow \infty}f(x_n)=l$$

means that there exists for all $\epsilon>0$ a $N_2$ such that for $n>N_2$

$$|f(x_n)-l|<\epsilon$$.

Why must these $N$ and $N_2$ exist? This is because of the definition of the limit of a sequence. Since $$\lim_{n\Rightarrow\infty} x=x_0 \Rightarrow \lim_{n\rightarrow \infty}f(x_n)=l,$$ we must have that $$|x_n-x_0|<\delta\Rightarrow |f(x_n)-l|<\epsilon.$$ There is nothing special with $x_n$. For all $x$ such that $|x-x_0|<\delta$ the above implication holds.

• I do understand the definition of a limit of a sequence quite well. It is also clear to me, that these two definitions of a function limit are in fact equivalent. What I am asking for is a Formal proof. What you have written is more of an informal explanation, why the sequential definition Implies the epsilon-delta one. What looks (to me) informal here? You have only considered one sequence and then tried to generalize it to all the numbers in given interval. – zoickx Oct 30 '16 at 22:37
• @IliaZaichuk, does the argumentation not hold for all sequences $\{x_n\}_{n=1}^\infty$ that converges to $l$? – FredikLAa Oct 30 '16 at 22:50