5
$\begingroup$

The informal intuition for the limit of a function is this:

What is the value of the function $f$ as $x$ gets infinitely close to $c$?

How on earth does this monster

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon > 0)(\exists \ \delta > 0) (\forall x \in D)(0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)$$ capture our intuition?

I want the answer to explain and motivate every bit of this formal statement. Remember: I am able to parse the statement, I just don't see how it captures our informal intuition.

$\endgroup$
  • 5
    $\begingroup$ Possible duplicate of Interpretation of $\epsilon$-$\delta$ limit definition $\endgroup$ – uniquesolution Oct 20 '15 at 21:10
  • $\begingroup$ This is a common problem when students first see $\epsilon-\delta$ definitions. They are a lot more subtle than they look. $\endgroup$ – Ali Caglayan Oct 20 '15 at 21:22
  • 1
    $\begingroup$ @uniquesolution The title may look similar, but the question is a different one: it's about the dependency of $\delta$ on $\epsilon$. The (only) answer to that question does not answer the present one. $\endgroup$ – user147263 Oct 20 '15 at 23:07
  • 1
    $\begingroup$ @uniquesolution I don't think the question is a duplicate. This question asks about the intuition behind the definition, while the linked question asks about how $\delta$ depends on $\epsilon$. In my opinion, this question is more fundamental (in the sense that someone first encountering limits will more likely ask this one, not the linked one). $\endgroup$ – 5xum Oct 20 '15 at 23:45
  • $\begingroup$ The informal definition of limit is not what you mention, bur rather somewhat more contrived. By $f(x) \to L$ as $x \to a$ we informally mean that "values of $f(x)$ can be made arbitrarily close to $L$ for all values of $x$ sufficiently close to (but not equal to) $a$". The phrases "arbitrarily close" and "sufficiently close" are informal and they are formalized by using $\epsilon - \delta$. An "informal definition" does not mean "non-rigorous definition", rather it means "avoiding the use of too many symbols". $\endgroup$ – Paramanand Singh Oct 21 '15 at 4:21
7
$\begingroup$

The definition capture not an open question as

''What is the value of the function $f$ as $x$ gets infinitely close to $c$''

but the exact statement:

$f$ can be as close as we want to $L$ if $x$ is sufficiently close to $c$.

If you think to this you can see that the ''monster'' works well.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This actually did answer my question. Thank you! $\endgroup$ – user132181 Oct 23 '15 at 11:20
20
$\begingroup$

The definition can be read, in human words:

For every positive $\epsilon$, there exists such a positive $\delta$ that if $|x-c|$ is smaller than $\delta$, but larger than $0$, then $|f(x) - L|$ is smaller than $\epsilon$.

First of all, lets's get the double inequality out of the way. Basically, $0<|x-c|$ is just saying that $|x-c|$ is not equal to $0$ (since it can't be negative), and saying that is just saying that $x-c$ cannot be $0$, or in other words, that $x$ is not allowed to equal $c$.

Now to a non-mathematician, that still makes very little sense, but take into account that $|a-b|$ is really the distance between numbers $a$ and $b$.

So, we can translate the definition into

For every positive $\epsilon$, there exists such a positive $\delta$ that if $x$ and $c$ are two distinct numbers and the distance between them is smaller than $\delta$, then the distance between $f(x)$ and $L$ is smaller than $\epsilon$.

But that still does not ring quite "natural" But what does the "for all $\epsilon$, there exists a $\delta$" in the beginning mean anyway? Well it means that whatever $\epsilon$ you give me, I can find a $\delta$ such that the condition will be true. So:

No matter what $\epsilon$ you choose, I can find such a positive $\delta$ that whenever you take any $x$ near (but not equal to) $c$ that is less than $\delta$ away from $c$ which are closer together than $\delta$, $f(x)$ will be closer than $\epsilon$ away from $L$.

Getting warmer to something readable? Well, let's get rid of the variables even further:

No matter how close you want $f(x)$ to be to $L$, I can tell you how close to $c$ you need to pick your $x$, and if you pick your $x$ that closely, then I can guarantee that $f(x)$ will be as close to $L$ as you originally wanted it to be.

This is very similar to what MPW wrote in comments:

$f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $c$

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ I think you've not only shown that the intuition is preserved, but that in fact the two statements are homotopic :) Very nice. $\endgroup$ – pjs36 Oct 20 '15 at 21:16
  • 1
    $\begingroup$ Nice, +1. One further step is usually something along the lines of "$f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $c$" $\endgroup$ – MPW Oct 20 '15 at 21:18
  • $\begingroup$ @MPW I added that into the answer. $\endgroup$ – 5xum Oct 20 '15 at 21:20
  • $\begingroup$ $f$ need not be defined at the point $c$, which is why the inequality posted by the OP is $0<|x-c|<\delta$, and not just $|x-c|<\delta$. Your first three yellow boxes actually require $f$ to be defined at $c$, which is just a tiny loss of generality. $\endgroup$ – uniquesolution Oct 20 '15 at 22:20
  • 1
    $\begingroup$ I'd really like to hear the explanation from the downvoters... what about the answer is bad enough to warrant a downvote? $\endgroup$ – 5xum Oct 23 '15 at 15:06
0
$\begingroup$

It means that we can make $f(x)$ take on a value as close as we like to $L$ (closer than any positive $\epsilon$ that we choose) by bringing $x$ near enough to $c$ (nearer than the positive $\delta$ which is asserted to exist in the definition).

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Imagine that you want to be $\epsilon$ close to to limit $L$, than the definition just says that you can pick arbitrary any $x$ from some $\delta$ neighborhood of $c$ and with $f(x)$ you are $\epsilon$ close.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Intuitively it makes sense if you read the right-hand side as saying "it is possible to get $f(x)$ as close to $L$ as you like by demanding only that $x$ get very near $c$". If the right-hand expression is always true, then the limit on the left will be true. This does not imply that $f(c)=L$, so $f$ is not necessarily continuous [that would be something like $\lim_{x->c}f(x)=f(c)$]. Finally, to gain an intuitive feeling for your monster take a function for which the left-hand limit fails [eg., $f(x)=0$ if $x<0$, $f(x)=1$ otherwise], then [with $c=0, L=1$] try to get the right-hand side to work [ie., given $\epsilon=0.1$ try to find a $\delta$ which works]. Do this and your monster will make sense.

| cite | improve this answer | |
$\endgroup$

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.