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For a general limit the $\epsilon-\delta$ definition of a limit (the formal definition of a limit) states that

$$\lim_{x \ \to \ a} f(x) = L \Leftrightarrow \forall \epsilon > 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies |f(x)-L| < \epsilon))$$

However the $\epsilon -\delta$ definition of a limit changes for limits as $x \to +\infty$ and $x \to -\infty$, and it changes again for limits that evaluate to $+\infty$ and $-\infty$


1. Limit as $x \to +\infty$

$$\lim_{x \ \to \ +\infty} f(x) = L \Leftrightarrow \forall \ \epsilon>0\; (\exists \ \delta : (\;x>\delta\implies |f(x) - L|\leq\epsilon)) $$

2. Limit as $x \to -\infty$

$$\lim_{x \ \to \ -\infty} f(x) = L \Leftrightarrow \forall \ \epsilon>0\; (\exists \ \delta : (\;x<\delta\implies |f(x) - L|\leq\epsilon)) $$

3. Limit evaluating to +$\infty$

$$\lim_{x \ \to \ a} f(x) = +\infty \Leftrightarrow \forall M > 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies f(x) >M)$$

4. Limit evaluating to -$\infty$

$$\lim_{x \ \to \ a} f(x) = -\infty \Leftrightarrow \forall N < 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies f(x) < N)$$


But why is that so? I understand that if you use the normal $\epsilon-\delta$ definition of a limit in these cases, you get contradictions that pop up like $0 < |x-\infty| < \delta \implies \delta > \infty$, which you cannot do anything further with.

While some of these differences might be subtle, it just seems counter-intuitive to change a formal and general definition.

I know that the fundamental idea behind the $\epsilon - \delta$ definition remains the same throughout all of these examples (that no matter how small we want to make our "error distance" ($\epsilon$) we can always find a "distance to our limit point" ($\delta$) that satisfies the definition of a limit) , but to get to that fundamental idea, the $\epsilon - \delta$ definition has to be subtly modified (and I'm not referring to modifications in notation) for each of these examples.

Or is it just a case that the $\epsilon - \delta$ definition for the general limit I put at the very start of this post is not as general as I thought?


Furthermore, how does the $\epsilon - \delta$ definition of a limit change for these cases, (the formal definitions of these don't seem to be covered in any introductory Calculus textbook).

5. Limit as $x \to +\infty$ $= +\infty$

$$ \lim_{x \ \to \ +\infty} f(x) = +\infty \Leftrightarrow \ ???$$

6. Limit as $x \to -\infty$ $= -\infty$

$$ \lim_{x \ \to \ -\infty} f(x) = -\infty \Leftrightarrow \ ???$$

7. Limit as $x \to -\infty$ $= +\infty$

$$ \lim_{x \ \to \ -\infty} f(x) = +\infty \Leftrightarrow \ ???$$

8. Limit as $x \to +\infty$ $= -\infty$

$$ \lim_{x \ \to \ +\infty} f(x) = -\infty \Leftrightarrow \ ???$$

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    $\begingroup$ How can your get within $\epsilon$ of infinity? $\endgroup$
    – John Douma
    Apr 20, 2016 at 21:52
  • $\begingroup$ @JohnDouma I see what you are saying, in that case are the last 4 cases I outlined not provable by the $\epsilon-\delta$ definition of a limit? $\endgroup$ Apr 20, 2016 at 21:55
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    $\begingroup$ Think of what the definitions are saying. If I approach a value I get a value. If that value is infinity, which is not a value, we have to say that we get arbitrarily large. Using your number 1 as an example, instead of saying if $x$ is within $\delta$ of infinity we say that if $x>\delta$ then $f(x)$ is within $\epsilon$ of $L$. $\endgroup$
    – John Douma
    Apr 20, 2016 at 21:59
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    $\begingroup$ The problem in that it is difficult to define infinity and infinitesimal. So, we come up with definitions that avoid these problematic concepts. Infinitesimal becomes $<\epsilon$ and infinite becomes > N. $\endgroup$
    – Doug M
    Apr 20, 2016 at 22:13
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    $\begingroup$ It's $\epsilon$-$\delta$, not $\epsilon - \delta$. Namely, it's a dash between $\epsilon$ and $\delta$, instead of a minus sign. $\endgroup$
    – Zhanxiong
    Jul 9, 2017 at 17:15

2 Answers 2

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We have to change the formal definition here because $\pm\infty$ are not members of $\mathbb{R}$.

One way to generalise this is to consider the set $ \mathbb{R}_\infty = \mathbb{R} \cup \{\infty\}$. Note here that we do not make a distinction between $\pm\infty$. We can define a Hausdorff topology on this set using the basis of open sets $$ \{U \subset \mathbb{R} : U \text{ is open}\} \cup \{B_\infty(\delta):\delta > 0\}$$ where we define the open ball about infinity as $B_\infty(\delta) = \{x \in \mathbb{R} : |x| > \delta\}$. This topological space is compact and called the one point compactification of $\mathbb{R}$. It is homeomorphic to a circle.

(If you're lost here because you haven't heard of topology before, just imagine wrapping the real line round a circle and gluing the ends together at a point which we call $\infty$. Then 'forget' what the original numbers were and treat all parts of the circle the same. In particular, limits work the same everywhere on this circle.)

You can prove that a function $ f : \mathbb{R}_\infty \to \mathbb{R}_\infty $ is continuous at a point $ a \in \mathbb{R} $ (using the standard definition for continuity in a Hausdorff space) if the restriction $f|_\mathbb{R}$ is continuous at $a$. Similarly you can prove that it is continuous at $a = \infty \in \mathbb{R}_\infty$ if the limits $\lim_{x\to+\infty} |f|_\mathbb{R}|$ and $\lim_{x\to-\infty} |f|_\mathbb{R}|$ exist (using your definitions for these limits, and the convention $\infty = |\pm\infty|$) and $$ \lim_{x\to+\infty} |f|_\mathbb{R}| = \lim_{x\to-\infty} |f|_\mathbb{R}| = f(\infty) .$$

It might please you to note (and prove) that, where it exists, $$ \lim_{x \to +\infty} f(x) = \lim_{x \to ^+0} f(1/x).$$ If you have a read about Möbius maps on $\mathbb{R}$, you will find that $x \mapsto 1/x$ is a Möbius map taking $0 \mapsto \infty$ and $\infty \mapsto 0$. This (algebraic) definition is inspired by the analysis here. The Möbius maps form a group under composition and are very important in several areas of mathematics (e.g. in the theory of Modular curves).


With regard to your definitions 5-8, you were very close with your previous guess. It is a useful exercise to practice writing down explicitly what you think you mean by phrases like "as $x \to \infty$".

When we talk about either $x$ or $f(x)$ tending to $\pm\infty$, we mean that it is 'getting really big' (in a positive or negative sense). We expressed 'getting really close' by using the $\epsilon$-$\delta$ definition you're happy with. In your definitions 1-4 you altered these to change one of the 'getting really close to' statements to a 'getting really big' statement.

To express $x \to +\infty$ in 1, you changed the relevant part of the statement to $(\exists \delta : x > \delta \Rightarrow \dots)$. To express $f(x) \to +\infty$ in 3, you changed the relevant part of the statement to $(\forall M > 0: (\exists \delta: \dots \Rightarrow f(x)>M))$.

Putting these characterisations together, we arrive at the definition for 5 $$\lim_{x\to+infty} f(x) = \infty \quad\Rightarrow\quad \forall M > 0 (\exists \delta: x > \delta \Rightarrow f(x) > M).$$

See if you can now do 6-8.

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To understand this, you need to think of the intuition behind the $\epsilon$-$\delta$ definition. We want $\lim_{x\to a}f(x)=L$ if we can make $f(x)$ as close to $L$ as we like by making $x$ sufficiently close to $a$. Worded differently, we might say that:

$\lim_{x\to a}f(x)=L$ if given any neighborhood $U$ of $L$, there is a neighborhood $V$ of $a$ such that elements of $V$ are mapped by $f$ to elements of $U$ (except possibly $a$ itself).

In this context, a "neighborhood" of a point $p$ should be understood to mean "points sufficiently close to $p$". Let's make that precise by defining what we mean by "close". For $\epsilon>0$ (assumed, but not required, to be very small) define $$B(x,\epsilon):=\{y\,:\,|x-y|<\epsilon\},$$ the ball of radius $\epsilon$ about $x$. For our purposes, we say $U$ is a neighborhood of $x$ if $U=B(x,\epsilon)$ for some $\epsilon>0$. (The usual definition only requires that $U$ contains such a ball.) Assuming $\epsilon>0$ is very small, this agrees with our intuition of what closeness should mean. Now if we go back to our neighborhood "definition" of a limit, you should be able to think about it for a bit and convince yourself that it is equivalent to the usual definition.

How does this relate to the problem with infinity? Given that infinity is not a real number (and things like distance from infinity do not make sense), we must revise what it means to be "close" to infinity. So for $M>0$ (assumed this time to be very large) define $$B(+\infty,M):=\{y\,:\,y>M\},\quad B(-\infty,M):=\{y\,:\,y<-M\},$$ the neighborhoods of $\pm\infty$. Hopefully you can see why these make sense as definitions; a number should be close to infinity if it is very large (with the correct sign), so a neighborhood of infinity should contain all sufficiently large numbers.

Now we extend our neighborhood definition of limits to include the case where $a$ or $L$ can be $\pm\infty$. It is a similar exercise to before to verify now that the definition is still equivalent to the old one, only now we have in some sense unified somewhat.

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  • $\begingroup$ just out of curiosity, where exactly did you get this definition $B(x, \epsilon) := \{y : |x-y| \leq \epsilon\}$, is it covered in Principles of Mathematical Analysis by any chance? $\endgroup$ Sep 9, 2016 at 16:56
  • $\begingroup$ This is a standard definition throughout analysis. $\endgroup$
    – user370967
    Jul 9, 2017 at 16:06

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