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Every time i get confused with the definition of $\lim_{x\rightarrow \infty}f(x)=L$. I could not find a reference that will give the definition.

I am trying to write what i understood. See if this is correct.

  • By $\lim_{x\rightarrow \infty}f(x)=L$ we mean the following : Given $\epsilon >0$ there exists $R>0$ such that $|f(x)-L|<\epsilon$ for all $x>R$.
  • By $\lim_{x\rightarrow -\infty}f(x)=L$ we mean the following : Given $\epsilon >0$ there exists $R<0$ such that $|f(x)-L|<\epsilon$ for all $x<R$.
  • By $\lim_{x\rightarrow \infty}f(x)=\infty$ we mean the following: Given $R>0$ there exists $L>0$ such that $|f(x)|>R$ for all $x>L$
  • By $\lim_{x\rightarrow \infty}f(x)=-\infty$ we mean the following: Given $R<0$ there exists $L>0$ such that $f(x)<R$ for all $x>L$
  • By $\lim_{x\rightarrow -\infty}f(x)=\infty$ we mean the following: Given $R>0$ there exists $L<0$ such that $|f(x)|>R$ for all $x<L$
  • By $\lim_{x\rightarrow \infty}f(x)=-\infty$ we mean the following: Given $R<0$ there exists $L<0$ such that $f(x)<R$ for all $x<L$.

Let me know if i understood somethings wrongly.

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  • $\begingroup$ It's quite fine for me. The signs for $R$ and $L$ are unncessary (they're implicit). $\endgroup$ – Bernard Apr 15 '16 at 14:25
  • $\begingroup$ You don't, technically, need $R<0//R>0$. for the first two. It can be any $R$. It's the "for all $x>R$" and "for all $x<R$" that makes the difference, not the sign of $R$. $\endgroup$ – Thomas Andrews Apr 15 '16 at 14:25
  • $\begingroup$ If there is no need for sign then $x\rightarrow \infty$ and $x\rightarrow -\infty$ would mean the same thing? @Bernard $\endgroup$ – user311526 Apr 15 '16 at 14:27
  • $\begingroup$ If there is no need for sign then $x\rightarrow \infty$ and $x\rightarrow -\infty$ would mean the same thing? @ThomasAndrews $\endgroup$ – user311526 Apr 15 '16 at 14:27
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    $\begingroup$ @topgeomj No, because one definition uses $x>R$ and the other $x<R$. That is the difference, not the sign of $R$. $\endgroup$ – Thomas Andrews Apr 15 '16 at 14:28
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I will use $+\infty$ in this answer to avoid ambiguity.

I would start with two definitions:

  • $\lim_{x\to +\infty} f(x) = L$ means for all $\epsilon>0$ there exists $M$ such that for all $x>M,$ $|f(x)-L|<\epsilon$ .

  • $\lim_{x\to+\infty} f(x)=+\infty$ means for all $R$ there exists $M$ such that for all $x>M$, $f(x)>R$.

Those two definitions let you define the other limits by symmetry:

$$\begin{align} \lim_{x\to+\infty} f(x) = -\infty&\iff \lim_{x\to+\infty} -f(x)=+\infty\\ \lim_{x\to-\infty} f(x) = M &\iff \lim_{x\to+\infty} f(-x)=M \end{align}$$

Where the $M$ in the second case can be any of either a real value, or $+\infty,-\infty$.

So $\lim_{x\to-\infty} f(x)= -\infty$ means $\lim_{x\to +\infty} -f(-x)=+\infty$, which means:

For any $R$ there is an $M$ so that for all $x>M$, $-f(-x)>R$.

Now, given any $R'$, you can set $R=-R'$ and find $M$ with this condition, and set $M'=-M$. Then if $x<M'$, $-x>M$, and thus $-f(x)>R$ or $f(x)<R'=-R$. So we get back the definition that we want.


The reason to distinguish $+\infty$ from $\infty$ is that some books use $\infty$ means a single point at infinity, in both directions - essentially, merging the two values $+\infty,-\infty$ into a single point at infinity.

Then:

$$\lim_{x\to\infty} f(x)= L \iff \lim_{x\to+\infty} f(x)=\lim_{x\to-\infty} f(x)=L$$

where $L$ can be any real or $+\infty,-\infty$.

$$\lim_{x\to W} f(x) = \infty\iff \lim_{x\to W} |f(x)|=+\infty$$

Where $W$ can be any real, or $+\infty$ or $-\infty$, or $\infty$.

Your question is somehwat confused, because you seem to distingush $\infty$ from $+\infty$ with absolute values when $\infty$ is the limit, but not when $x\to\infty$.

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  • $\begingroup$ Thanks... I understand this clearly... :) $\endgroup$ – user311526 Apr 15 '16 at 15:04
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One also does not need the absolute values on $|f(x)| > R$. If f is going to infinity, then $f(x) > R$ is accurate.

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  • $\begingroup$ For $f(x) = (-1)^{[x]}e^x$, $f(x) \to \infty | x \to +\infty$, but not $f(x) \to +\infty | x \to +\infty$. $\endgroup$ – Abstraction Apr 15 '16 at 14:34
  • $\begingroup$ Ok.. Can you say something about other definitions as well.. $\endgroup$ – user311526 Apr 15 '16 at 14:34
  • $\begingroup$ @Abstraction Some people distinguish between $+\infty$ and $\infty$, but quite often they are used interchangably. $\endgroup$ – Thomas Andrews Apr 15 '16 at 14:59
  • $\begingroup$ @ThomasAndrews Seeing as third and fifth examples in the initial post clearly distinguish between $f(x) \to +\infty$ and $f(x) \to \infty$ (using absolute value of $f(x)$), I think here the distinction is important. $\endgroup$ – Abstraction Apr 15 '16 at 15:08
  • $\begingroup$ Well, the main indication that the OP is using $\infty$ to mean a single point at infinity is the absolute values, but in the $x\to\infty$ examples, the definitions do not use $|x|>L$. So the clarity is not there in the question. @Abstraction $\endgroup$ – Thomas Andrews Apr 15 '16 at 15:11
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Let me know if i understood somethings wrongly.

You did. When you write $x \to \infty$ (as opposed to $x \to +\infty$), for $x \in \mathbb{R}$ it roughly means "when absolute value of $x$ is arbitrarily large" ($|x| > R$, not $x > R$)

There is a universal definition of limit using filters. A filter $\mathcal{F}$ is a collection of sets such that it doesn't contain an empty set and $\forall f_1, f_2 \in \mathcal{F}, \exists f_3 \in \mathcal{F} : f_3 \subseteq f_1 \cap f_2$. For example, a set of neighbourhoods of a real point $x$ is a filter ($\{(x-\epsilon, x+\epsilon) | \epsilon \in \mathbb{R}^+\}$, called neighbourhoods filter); a set of intervals with infinite endpoint ($\{(a,+\infty) | a \in \mathbb{R}\}$) is a filter; a set of segment complements ($\{(-\infty,a)\cup(b,+\infty) | a,b \in \mathbb{R}\}$) is a filter.

Now, let's take function $h \in \{X \to Y\}$ and there is a filter $\mathcal{F}$ on $X$ and a concept of "neighbourhoods" on $Y$ (called topology). Then let's take this sentence: "there is such $y \in Y$ that for any $y$ neighbourhood $O_y$, there is a filter element $f \in \mathcal{F}$ so that $h(f) \subseteq O_y$" and write it for short as $\lim_{\mathcal{F}}h(x) = y$.

Lets write filter $\{(a,+\infty) | a \in \mathbb{R}\}$ as $x \to +\infty$, filter $\{(-\infty,a) | a \in \mathbb{R}\}$ as $x \to -\infty$ and filter $\{(-\infty,a) \cup (a,+\infty) | a \in \mathbb{R}\}$ as $x \to \infty$.

Finally, let's call set $\{(L-\epsilon,L+\epsilon) | \epsilon \in \mathbb{R}\}$ a set of neighbourhoods of $L$, set $\{(b,+\infty) | b \in \mathbb{R}\}$ a set of neighbourhoods of $+\infty$, set $\{(-\infty,b) | b \in \mathbb{R}\}$ a set of neighbourhoods of $-\infty$ and set $\{(-\infty,b) \cup (b,+\infty) | b \in \mathbb{R}\}$ a set of neighbourhoods of $\infty$.

Now from these rules we can derive any definition. Take $\lim_{x \to -\infty}f(x) = +\infty$. It comes to "for any $+\infty$ neighbourhood $(b,+\infty), b \in \mathbb{R}$, there is a filter element $(-\infty,a), a \in \mathbb{R}$ so that $f((-\infty,a)) \subseteq (b,+\infty)$". Or, in more traditional notation, $$\forall b \in \mathbb{R}, \exists a \in \mathbb{R} : \forall x \in (-\infty,a), f(x) \in (b,+\infty)$$

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  • $\begingroup$ May be you are complication the question by saying more advanced things as filter, topology... $\endgroup$ – user312648 Apr 15 '16 at 16:09
  • $\begingroup$ @cello Maybe. But memorizing definitions for all cases of $x \to \pm\infty$ is actually hard, filters help to see some logic in all this. "Topology" here is simply a proper naming for a system of neighbourhoods; as you can see, it's properties aren't used. $\endgroup$ – Abstraction Apr 15 '16 at 16:21

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