One criterion for checking existence of limits is to check that and one-sided limits from left and right exist and agree:
(Theorem) Let $f$ be a real-valued function. One-sided limits of $f$ as $x$ approaches $a$ from the left and right exist and equal $L$:
$$
\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L
$$
if and only if the two-sided limit of $f$ at $a$ exists and also equals $L$,
$$\lim_{x \to a} f(x) = L.$$
The contrapositive of this statement can be used to conclude that a limit does not exist:
(Contrapositive) Let $f$ be a real-valued function. One-sided limits of $f$ as $x$ approaches $a$ from the left and right do not exist or do not agree if and only if the limit of $f$ at $a$ does not exist.
When applying this to a variety of contexts, you can come up with some pretty weird examples and weird results.
In the above figure, it seems to me that
- $\lim_{x \to 3} f(x)$ does not exist since $\lim_{x \to 3^+} f(x)$ does not exist.
- $\lim_{x \to 4} f(x)$ does not exist since $\lim_{x \to 4^-} f(x)$ and $\lim_{x \to 4^-} f(x)$ do not exist.
- $\lim_{x \to 5} f(x)$ does not exist since $\lim_{x \to 5^-} f(x)$ does not exist.
A bit more controversial is if you apply the same to the limits at infinity, which would stand to reason that
- $\lim_{x \to \infty} f(x)$ does not exist since $\lim_{x \to \infty^+} f(x)$ does not exist.
- $\lim_{x \to -\infty} f(x)$ does not exist since $\lim_{x \to \infty^-} f(x)$ does not exist.
However to contradict the above, many people would write $\lim_{x \to \infty} f(x)=2$ and $\lim_{x \to -\infty} f(x)=-1$.
TLDR; why does the nonexistence of left and right limits not cause limits at infinity to be undefined but does cause the limit at $x=3$, $x=4$, and $x=5$ to not exist? Real analysis answers are welcome.
Edit Thanks to answers from Troposphere and Joe, I worked out some more careful definitions and theorems:
Definition (Limit) Let $f$ be a function whose domain is a subset of $\mathbb{R}$, and let $a, L \in \mathbb{R}$. We say that the
limit of $f$ as $x$ approaches $a$ is $L$,
\begin{equation*} \lim_{x \to a} f(x)=L \end{equation*} to mean that $a$ is an accumulation point of $\textrm{dom}(f)$ and for any $\epsilon > 0$, there exists $\delta >0$ such that if $x \in \textrm{dom}(f)$ is within $\delta$ of $a$ (with $x \ne a$), then $f(x)$ is within $\epsilon$ of $L$: \begin{equation*} 0 < |x-a| < \delta \quad \rightarrow \quad 0 < |f(x)-L| < \epsilon.
\end{equation*}
Definition (One-Sided Limits) Let $f$ be a function whose domain is a subset of $\mathbb{R}$, and let $a, L \in \mathbb{R}$.:
We say the limit of $f$ as $x$ approaches $a$ from the left is $L$, $$\lim_{x \to a^-} f(x)=L,$$ to mean that $a$ is an accumulation point of $\textrm{dom}(f) \cap (-\infty,a]$ and for any $\epsilon > 0$, there exists $\delta >0$ such that if $x$ is within $\delta$ of $a$ (for $x < a$), then $f(x)$ is within $\epsilon$ of $L$: \begin{equation*} 0 < a-x < \delta \quad \rightarrow \quad 0 < |f(x)-L| < \epsilon. \end{equation*}
We say the limit of $f$ as $x$ approaches $a$ from the right is $L$, \begin{equation*} \lim_{x \to a^+} f(x)=L, \end{equation*} to mean that $a$ is an accumulation point of $\textrm{dom}(f) \cap [a,\infty)$ and for any $\epsilon > 0$, there exists $\delta >0$ such that if $x$ is within $\delta$ of $a$ (for $x > a$), then $f(x)$ is within $\epsilon$ of $L$: \begin{equation*} 0 < x-a < \delta \quad \rightarrow \quad 0 < |f(x)-L| < \epsilon. \end{equation*}
Theorem (One-Sided and Two-Sided Limits Relationship) Let $f$ be a function whose domain is a subset of $\mathbb{R}$, and let $a, L \in \mathbb{R}$. Suppose
a is an accumulation point of dom(f) andthat $f$ is defined everywhere in some punctured neighborhood of $a$. ThenOne sided limits of $f$ from left and right at $a$ exist and equal $L$, $$\lim_{x \to a^-} f(x) = L \textrm{ and }\lim_{x \to a^+} f(x) = L,$$
if and only if the two-sided limit of $f$ at $a$ exists and also equals $L$: $$\lim_{x \to a} f(x) = L.$$
Contrapositive (One-Sided and Two-Sided Limits Relationship) Let $f$ be a function whose domain is a subset of $\mathbb{R}$, and let $a, L \in \mathbb{R}$. Suppose
a is an accumulation point of dom(f) andthat $f$ is defined everywhere in some punctured neighborhood of $a$. ThenAt least one of the one sided limits of $f$ from left and right at $a$ does not exist or do not equal $L$: $$\lim_{x \to a^-} f(x) \ne L \textrm{ or }\lim_{x \to a^+} f(x) \ne L,$$
if and only if the two-sided limit of $f$ at $a$ does not exist or does not equal $L$: $$\lim_{x \to a} f(x) \ne L.$$
Under these definitions and theorems, I think (hope) we get the conclusions we expect:
- $\lim_{x \to 3} f(x)$ exists
- $\lim_{x \to 5} f(x)$ exists.
and
- $\lim_{x \to 3^+} f(x)$ DNE.
- $\lim_{x \to 4} f(x)$ DNE.
- $\lim_{x \to 5^-} f(x)$ DNE.
Best Answer
The general concept of "limit" is more fundamental than "one-sided limit", and the theorem you quote is specifically not a definition of $\lim_{x\to a}$.
In particular, the theorem strictly speaking has some hidden premises that are not clearly shown in the formulation you quote, namely that
Your conclusions don't work because the first hidden premise fails to hold in the $\infty$ cases, and the second hidden premise fails to hold in the $a\to3,4,5$ cases.
The general concept of limit can be defined* whenever the domain of the function is (a subset of) a topological space, and the $a$ in $x\to a$ is an accumulation point of the domain of the function in question.
For your example this means that limits as $x\to 3$ and $x\to 5$ make sense (and exist), but a limit as $x\to 4$ does not make sense, because $4$ is not an accumulation point of the domain. To speak about limits as $x\to\infty$ or $x\to-\infty$ we can take the underlying topological space to be the extended real line.
*namely: If $X$ and $Y$ are topological spaces, $Z$ is a (not necessarily proper) subset of $X$, $f$ is a function $Z\to Y$, $a$ is an accumulation point of $Z$, and $b\in Y$, we say that $b=\lim_{x\to a} f(x)$ iff:
(It's an instructive exercise to verify that this is equivalent to the usual $\varepsilon$-$\delta$ definition of limit in the special case where $X$ and $Y$ are metric spaces rather than merely topological spaces, and $Z$ includes a punctured neighborhood of $a$).
One practical reason to downplay the concept of one-sided limits is that it doesn't generalize well to higher dimensions. For example, if we take $$ f(x,y) = \frac{xy^2}{x^2+y^4} $$ defined on $\mathbb R^2\setminus\{(0,0)\}$, then $\lim_{h\to 0^+}f(hp,hq)=0$ for every $(p,q)\ne(0,0)$ -- that is, no matter which direction we approach $(0,0)$ from, the "single-directional limit" of $f$ exists and is $0$ -- yet $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist!