For a function $$f:X \rightarrow \mathbb{R}$$ My course notes says that:

$$\lim_{x \to \infty} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists S \in \mathbb{R}, x> S \implies |f(x)-l| < \epsilon)$$

I don't understand why we have dropped the $$\forall x \in X$$

In other words, why is it not this: $$\lim_{x \to \infty} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists S \in \mathbb{R}, \forall x \in X, x> S \implies |f(x)-l| < \epsilon)$$

Wouldn't we want the implication to hold true for all $x$ larger than $S$, analogous to when we're dealing with the definition of limits as $x$ approaches $a$?

$$\lim_{x \to a} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists \delta > 0, \forall x \in X, 0 < |x-a| < \delta \implies |f(x)-l| < \epsilon)$$

Otherwise, I could just find a really small $S$, smaller than a $x_1$ where $f(x_1)=l$ and if it works for one particular $x$, I could claim the limit as $x \rightarrow \infty$ is $l$, which is clearly not the intended definition of a limit as $x$ approaches infinity.

  • 2
    $\begingroup$ It should be. It wasn't written explicitly in the notes, but it should be there, just as you suspected. $\endgroup$ Nov 16, 2019 at 0:49
  • $\begingroup$ @bof Good catch. I edited the question a few times. Fixed now. $\endgroup$
    – Snowball
    Nov 16, 2019 at 1:12
  • $\begingroup$ " I could just find a really small S, smaller than a x1 where f(x1)=l and if it works for one particular x, " That makes no sense. The condition is $x > S \implies$ that means if any $x > S$ it is true. Not just one. Writing $\forall x \in X$ will have nothing to do with that. $\endgroup$
    – fleablood
    Nov 16, 2019 at 1:37

2 Answers 2


I will not enter into the details of first order logic, but strictly speaking, if a variable doesn't have a quantifier ($\exists$ or $\forall$) then it is ranging over all the elements of the set. So, again, strictly speaking both propositions are the same. In fact, strictly speaking, you can safely drop every $\forall x \in X$ whenever you know $X$ is the domain of discourse, it will just annoy your fellow mathematicians but it is right.

To see why this is the case, see this example (but remember that this is a formalism about mathematical language, so it is arbitrary in the sense that it was decided by convention to be so): if I am talking about the natural numbers and I say $x$ is even implies $x$ can be divided by $2$, one would reasonable understand that I meant, for all natural numbers $x$, $x$ is even implies it can be divided by 2. Clearly I wasn't talking about one specific $x$.

Again, just remember that this is a (intuitive) convention, nothing more than that.


The meaning of $x>S\implies |f(x)-l|<\epsilon$ is that "if $x>S$ then $|f(x)-l|<\epsilon$". So yes, it exactly says that for all $x$ which satisfy $x>S$ we have $|f(x)-l|<\epsilon$. An equivalent way to write the same thing:

$\forall (x: x>S)[|f(x)-l|<\epsilon]$

Note that in this equivalent way I didn't use $\implies$ at all.

  • $\begingroup$ Does this imply I can drop the $\forall x \in X$ in $\lim_{x \to a} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists \delta > 0, \forall x \in X, 0 < |x-a| < \delta \implies |f(x)-l| < \epsilon)$? $\endgroup$
    – Snowball
    Nov 16, 2019 at 0:53
  • $\begingroup$ If you know that you are checking the condition for elements of the set $X$ then yes, you can just write $0<|x-a|<\delta\implies |f(x)-l|<\epsilon$. Actually, I always write $\forall(x\in X: 0<|x-a|<\delta)[|f(x)-l|<\epsilon]$. But what I'm saying is that writing the way they did is also fine, as long as you know that you are checking the condition for elements from the set $X$. $\endgroup$
    – Mark
    Nov 16, 2019 at 0:59
  • $\begingroup$ This answer raises a question. According to this convention, WHERE within the expression should $\text{“ }\forall x \text{ ''}$ be? And why? $$ \begin{array}{ccccc} \text{here?} & & \text{or here?} & \text{or somewhere else?} \\ \downarrow & & \downarrow & \downarrow \\ \bullet\bullet\bullet & (\forall \varepsilon > 0 \,\,\, \exists \delta > 0 & \bullet\bullet\bullet & 0 < |x-a| < \delta \implies |f(x)-\ell| < \varepsilon) \end{array} $$ This matters, since in one case the $\delta$ depends on $x$ (which is NOT as it should be) and in the other it does not. $\qquad$ $\endgroup$ Nov 16, 2019 at 1:05
  • $\begingroup$ Obviously after the $\exists\delta>0$. $\endgroup$
    – Mark
    Nov 16, 2019 at 1:07
  • $\begingroup$ @Mark : That is obvious if you consider the meaning of the limit concept, but how does it follow from conventions of logic? $\endgroup$ Nov 16, 2019 at 1:08

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