This might be a silly question that might be trivial. I'm currently reviewing for my final, and going over a proof that says that:
Let $f:D \rightarrow R$ a function that is defined in a punctured neighbourhood of $x_0$, and assume $\lim_{x\rightarrow x_0}f(x)=L\in R$. There exists a punctured neighbourhood U of $x_0$ where $f$ is bounded.
What exactly is the significance of $f$ being defined in a punctured neighbourhood when it comes to limits?? What "qualities" or differences does a punctured neighbourhood have over a regular neighbourhood?
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$\begingroup$The function $\dfrac {\sin x} x$ is defined in the punctured neighbourhood $(-r, r) \setminus \{0\}$, because in $0$ it becomes $\dfrac 0 0$. Still, since $\lim \limits _{x \to 0} \dfrac {\sin x} x = 1$, this function may be extended by continuity in order to obtain a new one, continuous, this time defined everywhere: $f(x) = \begin {cases} \dfrac {\sin x} x, & x \ne 0 \\ 1, & x = 0 \end {cases}$.
Punctured neighbourhoods allow the variable of a function to get arbitrarily close to a "point that raises problems", while still not touching that point.
$\endgroup$ $\begingroup$A punctured neibourhood D of x in a topological space-also called a deleted $\epsilon$ neibourhood-can be visualized in the plane as the open disk D centered at x where x has been removed or "poked out".Imagine a sheet of paper represents a 2 dimensional space,D is any part of the sheet containing x which has an open disk containing x as a circle of radius $\epsilon$ that can be cut out with a scissors. Then a deleted $\epsilon$ neibourhood can be thought of as one of these circles where a pin of infinitesimal width has been used to poke out that point in the center so that any function defined on D isn't defined at x but is defined everywhere else.This gives a very simple and geometric interpretation of a limit; the function can get arbitrarily close to the "hole" in it's domain without actually being defined at the hole. Instead,it's defined everywhere else in the domain.
A simple example of such a function defined on a a deleted $\epsilon$ neibourhood is f:$R^2 \rightarrow R^2$ where f(x) = $\frac{1}{x}$. Clearly this function is a punctured nbh of 0 in the plane. Notice f(x) is defined on all other points in the plane and $\lim f(x)_{x\rightarrow 0}$ =$+\infty$.
$\endgroup$ $\begingroup$Topologically a neighbourhood is an open set. So a punctured neighbourhood $N$ of $x$, is $N=U\backslash \{x\}$, where $x_0\in U$, and $U$ is an open set, or open neighbourhood of $x$.
As an example, lets take our space as $\Bbb R$ and the set $D(a,r)=\{x\in \Bbb R\mid |x-a|<r\}$ is called a disk or radius $r$, centered at $a$. In this case $D(a,r)=(a-r,a+r)$. The punctured disk $D'(a,r)=D(a,r)\backslash \{a\}$ is a punctured neighbourhood of $a$. Sometimes this punctured disk is called $A(a,0,r)$ for an annulus, where $A(a,r_1,r_2)=\{x\in \Bbb R\mid r_1<\left|x-a\right|<r_2\}$.
As another example if you step the ideas in the last paragraph up a dimension to $\Bbb R^2$, and we measure distance the normal euclidean way, straight line distances, then the disks and annuli actually look like disks and annuli, and the punctured disk is like poking a hole through the centre of the disk.
These ideas of disks and annuli generalise to arbitrary sets with a notion of distance on them, called metric spaces, and metric spaces have a notion of open sets, like open intervals of $\Bbb R$.
The reason a punctured neighbourhood might be useful is that suppose $f$ is defined on $D'(a,r)$, the punctured disk as defined before, on a suitable space with distance, (like $\Bbb R$ or $\Bbb R^2$), is that $f$ can be defined "near" $a$ but need not be defined at $a$ at all. "Near" of course needs to be interpreted properly. Now supposing $f$ is defined this way, then even though $f$ has no value at $a$ we can consider what happens as we move closer to $a$.
This is useful as since $f$ is defined in a neighbourhood of $a$, and in $\Bbb R^n$ and $\Bbb C^n$ say, this will mean your $f$ is defined arbitrarily close to $a$, and so we can speak of $\lim\limits_{x\rightarrow a} \,f(x)$, or whether it exists or doesn't exist.
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