6
$\begingroup$

My Real Analysis course uses the $\epsilon - n_0$ definition of the limit, but I have noticed that the $\epsilon - \delta$ approach seems to be more common. Could someone please explain both formally and informally the difference between the definitions? Is there an advantage to using one over the other? An example of a simple proof using both definitions would be great!

(Sorry if this is a silly question. Looking at the definitions, I don't think that $n_0 = \delta$. but I could be wrong. If $n_0 = \delta$ then at least there is a simple answer, though I'll feel pretty embarrassed.)

$\endgroup$
  • $\begingroup$ I think you're mixing two different definitions of limit. The $n_0$ probably is the limit of a sequece while the $\delta$ one regards the limit of any real function. Do you mind writting both definitions here? $\endgroup$ – Git Gud Feb 18 '13 at 18:27
  • $\begingroup$ $\epsilon$-$n_0$ can be used to define the notion of limit of a sequence. $\epsilon$-$\delta$ can be just to define continuity of a function. $\endgroup$ – Hagen von Eitzen Feb 18 '13 at 18:27
  • 2
    $\begingroup$ You're right. I am confusing the definitions for functions and sequences. Though, the reason I got confused was that I was looking at the epsilon-delta proofs of the limit laws. I'll try doing the proofs of the limit laws for functions and sequences side by side and re-ask about that specifically if I get stuck. Thanks! $\endgroup$ – economista Feb 18 '13 at 18:38
4
$\begingroup$

A sequence $(x_n)_{n\in\mathbb N}$ of real numbers has limit $x$ iff for every $\epsilon>0$ there exists $n_0\in\mathbb N$ such that $n>n_0$ implies $|x_n-x|<\epsilon$.

A function $f\colon I\to \mathbb R$ is continuos at $x_0\in I$ if for every $\epsilon>0$ there exists a$\delta>0$ such that for all $x\in I$ with $|x-x_0|<\delta$ we have $|f(x)-f(x_0)|<\epsilon$.

These are different concepts. Then again, they are the same: Note that $|x-x_0|$ measures the distance between $x$ and $x_0$; we can define a metric $d$ on $\mathbb N\cup\{\infty\}$ such that a sequence $(x_n)_{n\in\mathbb N}$ converges to $x$ if and only if the function given by $$f(n)=\begin{cases}x_n&n\in\mathbb N\\x&x=\infty\end{cases}$$ is continuous at $\infty$. The metric I mentioned can be defined as $d(n,m)=\left|\frac 1n-\frac 1m\right|$ for $n,m\in\mathbb N$ and $d(n,\infty)=d(\infty,n)=\frac 1n$ and $d(\infty,\infty)=0$.

$\endgroup$
  • 1
    $\begingroup$ All good answers, but I appreciate your thoroughness on this basic question. $\endgroup$ – economista Feb 18 '13 at 18:39
  • 1
    $\begingroup$ No need to talk about continuity, just that $\lim_{x \rightarrow x_0} f(x) = a$ if ... If now $\lim_{x \rightarrow x_0} f(x) = f(x_0)$, then $f$ is continuous at $x = x_0$. $\endgroup$ – vonbrand Feb 19 '13 at 1:49
3
$\begingroup$

There is a difference $\epsilon-n_0$ is for the convergence of sequences, while $\epsilon-\delta$ is of continuity of functions.

In the following I will use the definition with the metric induced by the absolute value.

The $\epsilon - n_0$ definition is assumed to be: Let $(a_n)_{n\in\mathbb{N}}$ be a sequence, we call $(a_n)_{n\in\mathbb{N}}$ convergent if there is an $a$ so that $$\forall \epsilon >0 \ \exists n_0 \in \mathbb{N}:\ n>n_0\implies \ |a_n-a|< \epsilon $$

The $\epsilon-\delta$ definition is assumed to be. Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function, we call $f$ continuous at $x_0$ if $$\forall \epsilon >0 \ \exists \delta>0 \ : | x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon$$

$\endgroup$
1
$\begingroup$

The $n_0$ is (as noted) for a limit of a sequence, or sometimes for a limit as the variable goes to infinity, as in $$ \lim_{x\to\infty}\frac{1}{x} = 0 $$ phrased as: for every $\epsilon > 0$ there is $n_0 \in \mathbb R$ such that for all $x > n_0$ we have $$ \left|\frac{1}{x} - 0\right| < \epsilon . $$

$\endgroup$
0
$\begingroup$

I would say $\epsilon-n_0$ is more commonly used when you want the limit of a sequence $\left \{a_n \right\}$ i.e. $\forall n>n_0, |c-a_n|<\epsilon$. But $\epsilon-\delta$ sounds like $\forall \epsilon>0, \exists\delta>0$ such that $\forall x, |f(x+\delta)-f(x)|<\epsilon$ when you try to check the continuity of a function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.