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. $\endgroup$ – Andrés E. Caicedo Nov 16 '19 at 0:49
  • $\begingroup$ @bof Good catch. I edited the question a few times. Fixed now. $\endgroup$ – Snowball Nov 16 '19 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 '19 at 1:37

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 '19 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 '19 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. $\qquad$ $\endgroup$ – Michael Hardy Nov 16 '19 at 1:05
  • $\begingroup$ Obviously after the $\exists\delta>0$. $\endgroup$ – Mark Nov 16 '19 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? $\endgroup$ – Michael Hardy Nov 16 '19 at 1:08

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.