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The definition of the limit states that limit of $f(x)$ when $x$ approaches $c$ is $L$ iff for every $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - L | < \epsilon$ and $0 < |x - c| < δ )$.

This states that $f(x)$ can reach $L ( L- ε < f(x) < L + \epsilon)$, while $x$ cannot reach $c ( 0 < |x-c|)$.

The informal definition says that limit means the value the function approaches as the input approaches some value. (They use the same word.) Why can one reach its corespondent value ($L$) while the other can't ($x$ to equal $c$)? Why f(x) can equal L , and x can't equal c. What is the intuitive answer to this question?

Lazu Razvan
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  • Short answer is it's because we want the possibility for a limit to exist at a point $x=c$ even if the function isn't defined at $x=c$. Saying $0 < |x-c|$ allows the function to not be defined at $x=c$. –  Sep 27 '19 at 12:12

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The definition is not quite what you wrote. It should be:

For all $\varepsilon > 0$, there exists $\delta > 0$ such that, if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$.

Putting the $0 < $ next to the $|x - c| < \delta$ actually makes the definition weaker, in the sense that it's easier to satisfy. By putting the $0 < $ there, it means that we don't care what happens when $x = c$. The if...then statement imposes something to be true whenever $0 < |x - c| < \delta$ is true. When $0 < |x - c| < \delta$ is false (e.g. if $x = c$), the premise is false, so the conclusion $|f(x) - L| < \varepsilon$ can be true or false.

If we were to get rid of the $0 < $, we actually recover the concept of continuity, i.e. we don't just get a limit, we actually force the limit $L$ to be $f(c)$. Why? Because we can always choose $x = c$, and knowing that $$|x - c| < \delta \implies |f(x) - L| < \varepsilon$$ implies that, in the case of $x = c$, $$0 < \delta \implies |f(c) - L| < \varepsilon.$$ Note that $0 < \delta$ is always true, hence the conclusion $|f(c) - L| < \varepsilon$ must also be always true, regardless of $\varepsilon$. But, there's only one number $L$ so that $|f(c) - L| < \varepsilon$ for all $\varepsilon > 0$, and that's $L = f(c)$.

On the other hand, if we added the $0 <$ to the conclusion $|f(x) - L| < \varepsilon$, this becomes a stronger definition, in the sense that it's harder to satisfy. Not only do we require that, on the interval $(c - \delta, c + \delta)$, the function to be $\varepsilon$-close to $c$, we also require it to be different from $c$. This would exclude constant functions from having limits, as on every $\delta$-neighbourhood, they take their limit as a function value at every point!

Theo Bendit
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    "When 0<|x−c|<δ is false (e.g. if x=c), the premise is false, so the conclusion |f(x)−L|<ε can be true or false." This illustrates a common misconception well, that I think the OP might have made. There's nothing stopping a function being equal to L anywhere, even at more places than just c. The idea that an asymptote (effectively a type of limit) can "never reach the thing it's asymptotic to" is just plain wrong – Jam Sep 27 '19 at 12:33
  • what do you mean by "recover the concept of continuity" ? – Lazu Razvan Sep 28 '19 at 23:01
  • @LazuRazvan I just mean that, if you modify the definition of limits to remove the $0 <$, then the result is equivalent to continuity of the given function at the given point. – Theo Bendit Sep 29 '19 at 00:30
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The idea is that if, for instance,$$f(x)=\begin{cases}1&\text{ if }x=0\\2&\text{ otherwise,}\end{cases}$$then $\lim_{x\to0}f(x)=2$. The fact that $f(0)=1$ is not relevant here. What matters is that when $x$ is close to $0$ and distinct from $0$, then $f(x)$ is close to $2$. And in this case it is actually equal to $2$. So, if we had imposed that $0<\bigl\lvert f(x)-L\bigr\rvert$, the limit would not exist.

José Carlos Santos
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  • While this is true ,this is not quite what i am looking for .This is a specific case of limit use in which the definition help us calculate the function you wrote. I want to find a way to understand the meaning of the limit in general , and why that specific formal definition describes that meaning perfectly .I want to transcribe that defintion into something intuitive – Lazu Razvan Sep 27 '19 at 14:11
  • @LazuRazvan This is the reason why. If we add the additional condition that $f(c)$ equals the limit, we get continuity. Thus, we can have a discontinuous function in which the limit still exists. – John Douma Apr 25 '21 at 17:31
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Indeed by the definition of limit

$$\lim_{x\rightarrow c} f(x) = L \iff (\forall \varepsilon >0\, \exists \delta > 0: \forall x\in D\quad \color{red}{0<\vert x-c\vert <\delta} \implies \vert f(x)-L\vert <\varepsilon $$

it suffices that $f(x)$ approches $L$ as $x$ approaches $c$, as for example

$$\lim_{x\to 0} \frac{\sin x}{x}=1$$

for which we can define any value for $f(0)$ or also assume $f(x)$ not defined at $x=0$.

user
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  • it seems like this is more an advantage / a positive side effect of the limit , than the underlying meaning of the definition of the limit . – Lazu Razvan Sep 27 '19 at 14:06
  • @LazuRazvan It is exactly the aim in the definition of limit, that is $\forall x$ s.t. $0<|x-c|<\delta$ ($\implies x\neq c$) $\cdots$ and so on. We don't care of $f(c)$. – user Sep 27 '19 at 14:10
  • Ok,i think i might have overthought it , but then why f(x) can be L (can be equal to L) while x can't reach c – Lazu Razvan Sep 28 '19 at 12:22
  • @LazuRazvan The fact that $f(c)=L$ or $f(c)\neq L$ or that $f(c)$ is undefined is really not important for the definition of limit. Of course it can be $f(c)=L$ but it is inessential. – user Sep 28 '19 at 12:33