2
$\begingroup$

Is $\large{\lim_{x \to a}f(x)=L}$ The same as saying "one can get $\large{f(x)}$ as close as imaginable to $\large{L}$, by setting $\large{x}$ close enough to $\large{a}$" and vice versa "one can get $\large{x}$ as imaginable to $\large{a}$ by setting $\large{f(x)}$ close enough to $\large{L}$"?

Sorry if obvious questions, but I need to understand this before i can grasp the epsilon delta definition of a limit.

$\endgroup$
5
  • $\begingroup$ I don't think "as close as you like by setting $x$ close enough to $a$" really captures $f(x) \to L$, because that to me sounds like "for each $\epsilon>0$ there's some $x$ close to $a$ such that $f(x)$ is within $\epsilon$ of $L$". But you need $f(x)$ close to $L$ on a neighbourhood of $a$ for the definition of $f(x)\to L$ to hold. $\endgroup$
    – Matthew Towers
    May 25, 2015 at 8:15
  • $\begingroup$ But does it not follow, for example in the case of $\large{\lim_{x \to a}f(x)=L}$, that if you can get f(x) to +-$\epsilon$ of L by setting x close enough to a, then if you set x closer to a, every function evaluated at those x'es will be within $\epsilon$ of f(x). So what i'm basically trying to ask is if it's not wrong, to define the limit the way i did. $\endgroup$
    – Carefullcars
    May 25, 2015 at 8:47
  • 1
    $\begingroup$ No, that doesn't follow. Say $x<y<a$. It's possible that $f(x)$ is within $\epsilon$ of $L$ but $f(y)$ isn't, even in the case $f(x) \to L$. $\endgroup$
    – Matthew Towers
    May 25, 2015 at 11:55
  • 1
    $\begingroup$ hmm, yes i think i see what you are saying. You have to say(given that a limit exists) that for every possible $\epsilon$ range around L there exists some range around x, such that every x within that range evaluates to f(x) that is within $\epsilon$ . Right? $\endgroup$
    – Carefullcars
    May 25, 2015 at 12:13
  • $\begingroup$ Yes, that's it exactly. $\endgroup$
    – Matthew Towers
    May 25, 2015 at 12:17

1 Answer 1

Reset to default
5
$\begingroup$

In your case, roughly speaking you could say "one can get $f(x)$ as close as imaginable to $L$, by setting $x$ close enough to $a$". About the second statement, it is not right, since there may be values of $x$ for which $f(x)=L$ but $x\neq a$, or $x$ is "very far" from $a$. For example, think of the constant function $f(x)=L$. Then, $$ f(0)=L = \lim_{x\to 50} f(x). $$

Here $a=50$, but $f(x)$ approaching $L$ does not mean $x$ approaches $50$; this is more related to injectivity of the function.

$\endgroup$
0

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.