# Rephrasing the definition of a limit

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.

• 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. May 25, 2015 at 8:15
• 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. May 25, 2015 at 8:47
• 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 $f(x) \to L$. May 25, 2015 at 11:55
• 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? May 25, 2015 at 12:13
• Yes, that's it exactly. May 25, 2015 at 12:17

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.