Consider $f(x)$ function . We want to calculate $\lim_{x \to 3}f(x)$. So for left limit , we approach to $3$ and then compute $f(2.9) , f(2.99) , f(2.999)$ and so on . Now there is a weird thing . It is obvious that $2.9999.... = 3$ and also when we are talking about limit , point isn't important . In this case we don't take care about $f(3)$ but when we approach to $3$ infinitely , we get $3$ as $2.9999.... = 3$ ! . I'm very confused about these two concepts .

  • $\begingroup$ See here. $\endgroup$ – Dietrich Burde Apr 7 '17 at 15:20
  • 4
    $\begingroup$ But what is the question ? $\endgroup$ – A---B Apr 7 '17 at 15:39
  • 3
    $\begingroup$ As I understand this it's not really about why $2.999\ldots=3$. There are plenty of $0.999\ldots=1$ questions on this site already with great answers anyways. I think this is about how the limit of $f(x)$ as $x$ approaches $3$ from below along $2,2.9,2.99,\ldots$ can be different from the function value at $2.999\ldots=3$. Is that what your question is about, or is it something else? $\endgroup$ – Arthur Apr 7 '17 at 16:58
  • 1
    $\begingroup$ @Arthur Yes , that is . $\endgroup$ – S.H.W Apr 7 '17 at 17:03
  • 1
    $\begingroup$ Related Mathematics SE questions : Is any real-valued function in physics somehow continuous? , Computability, Continuity and Constructivism . Bottom line: it's an intelligent question and the OP is not the only one who is struggling with these issues. $\endgroup$ – Han de Bruijn Apr 12 '17 at 11:42

No, in order to find that some real $l$ is the limit $$ \lim_{x\to3^-}f(x) $$ you don't compute $f(2.9)$, $f(2.99)$ and so on. And neither you compute $f(2.(9))$ (periodic $9$), because no assumption is made that $f$ is defined at $3$, nor the possible value of $f$ at $3$ is relevant for the existence of the limit.

Saying that $$ \lim_{x\to3^-}f(x)=l $$ means

for every $\varepsilon>0$ there exists $\delta>0$ such that, for $3-\delta<x<3$, it holds $|f(x)-l|<\varepsilon$.

You can compute $f(2.9...9)$, if you wish; it may give you an idea of what $l$ could be, but in general it won't.

  • $\begingroup$ I'm sorry but I'm still confused . My problem is this : What's the difference between closing to a number infinitely and reaching it ? $\endgroup$ – S.H.W Apr 9 '17 at 16:07
  • 1
    $\begingroup$ @S.H.W There is no “closing to a number” in the definition I gave. Some teachers still like to use that language, notwithstanding that precise and rigorous definitions have been developed about 200 years ago. $\endgroup$ – egreg Apr 9 '17 at 16:16
  • $\begingroup$ Please take a look at it : mathsisfun.com/calculus/limits-formal.html $\endgroup$ – S.H.W Apr 9 '17 at 16:20
  • $\begingroup$ I think as you can see in "Mathisfun" page , the epsilon-delta definition is same as closing to a number . $\endgroup$ – S.H.W Apr 9 '17 at 16:32
  • 2
    $\begingroup$ @S.H.W I recommend you not to learn math from that site. $\endgroup$ – egreg Apr 9 '17 at 18:03

The idea of using a limit $\mathit{x}\rightarrow \mathit{n}$ is that you approach to $\mathit{n}$ as close as possible, but you actually never reach it.

Just forget that you are "computing" $f$ at every point because it is a missunderstanding. Imagine that you are moving along the graph of the function $f$, then when taking a limit you are getting as close as possible to a specific point without ever touching it, as the function does not need to be $defined$ at that point, or the image might be different than the limit itself.

Imagine the following case:

$$f(x) = \left \lbrace {x^2, x \not= 0 \atop 1 , x = 0}\right. $$

If you take $lim_{x\rightarrow0}f(x) = 0$ for both right and left limits, but the actual image is $f(0)=1$.

When one has the equallity between right limit, left limit and image at a certain point in a function, we then say that the function is $continuous$, but any function that is not continuous still has limits.

I hope I clarified that to you.

edit: keep in mind that when talking about real numbers, between any two numbers there is an infinity of more numbers, it doesn't matter how close you try to imagine them to be, and that is the idea exploited by the limit.

  • $\begingroup$ I'm sorry but I'm still confused . My problem is this : What's the difference between closing to a number infinitely and reaching it ? $\endgroup$ – S.H.W Apr 9 '17 at 16:06
  • $\begingroup$ @S.H.W Ayeron gives a good example of the difference. As long as $x$ is not $0$ - no matter how close to $0$ it is then $f(x)=x^2$. But when $x=0$ ("reaching it" in your terms) $f(x)=1$. BTW, in standard real analysis $\epsilon, \delta$ are real numbers. Being real numbers they are not infinite, infinitesimals, or infinitely close to other real numbers. In other words there is a distinction between "arbitrarily close" and "infinitely close" or similar statements. There is a branch of math called "non-standard real analysis" that deals with infinitesimals. Wikipedia has an article on it. $\endgroup$ – Χpẘ Apr 17 '17 at 1:36
  • $\begingroup$ @user2460798 Consider this definition : $\forall \epsilon > 0 \; \exists \; \delta > 0: |x- 3| < \delta \Rightarrow |f(x) - L | < \epsilon$ . Can we choose $2.999...$ for $x$ ? $\endgroup$ – S.H.W Apr 17 '17 at 13:52
  • $\begingroup$ @user2460798 You said , "there is a distinction between "arbitrarily close" and "infinitely close" or similar statements." . In real analysis for finding the limit we are dealing with infinitely close or arbitrarily close to a number ? $\endgroup$ – S.H.W Apr 17 '17 at 13:57
  • $\begingroup$ Arbitrarily close. $\endgroup$ – Χpẘ Apr 17 '17 at 15:50

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.