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Can someone tell me if I'm thinking about this definition correctly? I tried to summarize the precise definition of a limit by saying:

$F(x)=L$ over some range of $y$ values correlates to the range of possible $x$ values that make a limit statement plausible.

So as the range of $f(x)$ values get smaller and closer to $L$ , the range of $x$ values also get smaller and closer to $a$. And the same situation if one range increases, the other range also increases.

Even if I am on the right track, can someone explain it in different way so that I may better understand this?

Juniven Acapulco
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user21718
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  • maybe you need this http://math.stackexchange.com/questions/945755/understanding-of-the-formal-and-intuitive-definition-of-a-limit?rq=1 – Juniven Acapulco Dec 27 '16 at 07:17
  • Basically you are given $|f(x) - L|$, we want this to be as small as possible. So I give you any number say $\epsilon > 0$ and bound $|f(x) - L| < \epsilon$, to make this as small as possible, we approach $x \to a$. When $|x - a|$ is small enough, say for some number $\delta > 0$, we get the desired bound on $|f(x) - L| < \epsilon$. The idea now is to give the smallest possible $\epsilon > 0$. I don't know if this answers your question or not. – IAmNoOne Dec 27 '16 at 07:22

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So as the range of f(x) values get smaller and closer to L , the range of x values also get smaller and closer to a

No offense, but that is a pretty confused sentence, if you ask me.

I don't understand what exactly you mean by most of the words you use, so I will rather write my own explanation.

The definition of a limit is that

$\lim_{x\to a} f(x)=L$ is true if and only if, for every $\epsilon>0$, there exists $\delta>0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.

This definition tells you that, no matter what $\epsilon$ I pick, you can find me a $\delta$ such that $f(x)$ will be $\epsilon$ - close to $L$ if $x$ is $\delta$-close to $a$.

With fewere variables, this would translate to

I can get $f(x)$ to be as close to $L$ as I want, if I only put $x$ close enough to $a$.

5xum
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