Mathematical Analysis
Lecture 7
4.1 Cluster points
Definition 4.1.1. Let $S \subset \R$ be a set. A number $x \in \R$ is called an cluster point of $S$ if for every $\epsilon \gt 0$, the set $(x-\epsilon,x+\epsilon) \cap S \setminus \{ x \}$ is not empty.
👉 cluster point = accumulation point
Example 4.1.1.
$\left\{\dfrac{1}{n}~:~ n \in \N\right\}$
$0$ is the only cluster point.
Example 4.1.1.
$(0,1) = \left\{x\in \R~:~ 0\lt x\lt 1 \right\}$
The cluster points are all points in the closed interval $\left[0,1\right]$.
Example 4.1.1.
$\left[0,1\right)\cup\left\{2\right\}=\{x\in\mathbb R: 0\leq x\lt 1\}\cup\left\{2\right\}$
The set of cluster points is the interval $\left[0,1\right]$.
Example 4.1.1.
No cluster points.
Example 4.1.1.
The set of cluster points is $\R$.
4.1 Cluster points
A number $c \in \R$ is a cluster point of $S\subset \R$
if and only if
there exists a convergent sequence of numbers $\{ x_n \}_{n=1}^\infty$ such that $x_n \not= c$ and $x_n \in S$ for all $n$,
and $\lim\limits_{n\to\infty} x_n = c$.
4.1 Cluster points
Theorem 4.1.1. A number $c \in \R$ is a cluster point of $S\subset $ if and only if there exists a convergent sequence of numbers $\{ x_n \}_{n=1}^\infty$ such that $x_n \not= c$ and $x_n \in S$ for all $n$, and $\lim\limits_{n\to\infty} x_n = c$.
Proof. $\nec$ Suppose $c$ is a cluster point of $S$. For every $n \in \N$, pick $x_n$ to be an arbitrary point of the set $$\ds\left(c-\frac{1}{n},c+\frac{1}{n}\right) \cap S \setminus \{c\}$$ This set is not empty since $c$ is a cluster point of $S$. Then $\ds \abs{c-x_n} \lt \frac{1}{n} .$ Since $\left\{ \dfrac{1}{n} \right\}_{n=1}^\infty\to 0,$ $\{ x_n \}_{n=1}^\infty \to c. \; \bs$
4.1 Cluster points
Theorem 4.1.1. A number $c \in \R$ is a cluster point of $S\subset $ if and only if there exists a convergent sequence of numbers $\{ x_n \}_{n=1}^\infty$ such that $x_n \not= c$ and $x_n \in S$ for all $n$, and $\lim\limits_{n\to\infty} x_n = c$.
Proof. $\suf$ Now assume that a sequence of numbers $\{ x_n \}_{n=1}^\infty$ in $S$ converges to $c$ such that $x_n \not= c$ for all $n$, then for every $\epsilon \gt 0$ there is an $N$ such that for all $n\geq N$ we have $\abs{x_n - c} \lt \epsilon.$ In particular \[ \abs{x_N - c} \lt \epsilon. \] That is, $x_N \in (c-\epsilon,c+\epsilon) \cap S \setminus \{c\}. \; \bs$
4.1 Limits of functions
Definition 4.2.1. Let $f \colon S \to \R$ be a function and $c$ an cluster point of $S \subset \R$. Suppose there exists an $L \in \R$ and for every $\epsilon > 0$, there exists a $\delta \gt 0$ such that whenever $x \in S \setminus \{ c \}$ and $\abs{x - c} \lt \delta$, we have \begin{equation*} \abs{\,f(x) - L} \lt \epsilon . \end{equation*} We then say $f(x)$ converges to $L$ as $x$ goes to $c$.
4.1 Limits of functions
Definition 4.2.1. Let $f \colon S \to \R$ be a function and $c$ an cluster point of $S \subset \R$. Suppose there exists an $L \in \R$ and for every $\epsilon > 0$, there exists a $\delta \gt 0$ such that whenever $x \in S \setminus \{ c \}$ and $\abs{x - c} \lt \delta$, we have \begin{equation*} \abs{\,f(x) - L} \lt \epsilon . \end{equation*} We then say $f(x)$ converges to $L$ as $x$ goes to $c$.
We say $L$ is the limit of $f(x)$ as $x$ goes to $c$. We write \begin{equation*} \lim_{x \to c} f(x) := L , \end{equation*} or $f(x) \to L$ as $x \to c .$ If no such $L$ exists, then we say that the limit does not exist or that $f$ diverges at $c$.
4.1 Limits of functions
Geometric interpretation
Example 4.2.1.
Claim: $\displaystyle \lim_{x\ra 2} 3x+1$ $= 7.$
Discussion: For every $\vre >0 $, we need to find a $\delta>0$ so that \[ \abs{x-2}\lt\delta \;\; \implies \;\; \abs{(3x+1) - 7}\lt \vre. \] So we start from what we want to estimate:
$\abs{(3x+1) - 7} $ $ = \abs{3x-6} $ $ = 3\abs{x-2}$ $ \lt 3\delta.$
Example 4.2.1.
Claim: $\displaystyle \lim_{x\ra 2} 3x+1 = 7$.
Discussion: If we choose $\delta := \dfrac{\vre}{3}$, then $$\abs{(3x+1) - 7} \lt \vre .$$ Now we are ready to write a formal proof. 😃
Example 4.2.1.
Claim: $\displaystyle \lim_{x\ra 2} 3x+1 = 7$.
Proof. Let $\vre >0 $. Choose $\delta := \dfrac{\vre}{3}.$ Thus
$0\lt \abs{x-2}\lt \delta$
$ \implies \abs{(3x+1) - 7}$
$ \lt 3 \left(\dfrac{\vre}{3} \right)$
$ = \vre.$
$ \;\blacksquare$
Example 4.2.2.
Claim: $\displaystyle \lim_{x\ra 2} x^2 $ $=4.$
Discussion: Given an arbitrary $\vre >0 $, our goal is to find a $\delta>0$ so that \[ \abs{x-2}\lt\delta \; \implies \; \abs{x^2 - 4}\lt \vre. \] Note that $\abs{x^2 - 4} = \abs{x+2}\abs{x-2} .$ We need to manipulate the inequality $\abs{x-2}\lt\delta$ in order to get an upper bound for $\abs{x+2}.$
Example 4.2.2.
Claim: $\displaystyle \lim_{x\ra 2} x^2 = 4$.
Discussion: That is
$\abs{x-2} \lt \delta$
$-\delta \lt x-2 \lt \delta \quad\;\;$
$4 -\delta \lt x + 2 \lt 4 + \delta \quad$
$ \quad\;\; \abs{x+2} \lt \delta + 4$
If $\delta =1$, then $\abs{x+2} \lt 1+ 4 $ $ = 5. $
Thus, if we choose $\delta := \min\{1, \vre / 5\}$, $\abs{x-2}\lt\delta$ implies $\abs{x^2 - 4} \lt \vre $.
Example 4.2.2.
Claim: $\displaystyle \lim_{x\ra 2} x^2 = 4$.
Proof.
Let $\vre >0 $.
Choose $\delta := \min\left\{1, \dfrac{\vre }{5}\right\}$.
For all $x\in \R \setminus \{2\}$ and $\abs{x-2}\lt\delta$,
we have that
$\abs{x^2-4} $ $ \lt \abs{x+2}\abs{x-2}$ $\lt (5) \dfrac{\vre}{5}$ $\lt\vre.$ $\; \blacksquare$
4.1 Limits of functions
Theorem 4.2.1. Let $c$ be an cluster point of $S \subset \R$ and let $f \colon S \to \R$ be a function such that $f(x)$ converges as $x$ goes to $c$. Then the limit of $f(x)$ as $x$ goes to $c$ is unique.
Proof. Let $L_1$ and $L_2$ be two numbers that both satisfy the definition. Take an $\vre > 0$ and find a $\delta_1 > 0$ such that $\abs{f(x)-L_1} \lt \dfrac{\vre}{2}$ for all $x \in S \setminus \{c\}$ with $\abs{x-c} \lt \delta_1$.
Also find $\delta_2 > 0$ such that $\abs{f(x)-L_2} \lt \dfrac{\vre}{2}$ for all $x \in S \setminus \{c\}$ with $\abs{x-c} \lt\delta_2$.
4.1 Limits of functions
for all $x \in S \setminus \{c\}$ $\abs{x-c} \lt\delta_1 \,\Ra \,\abs{f(x)-L_1} \lt \dfrac{\vre}{2}$
for all $x \in S \setminus \{c\}$ $\abs{x-c} \lt\delta_2 \,\Ra \,\abs{f(x)-L_2} \lt \dfrac{\vre}{2}$
Proof. Now put $\delta := \min \{ \delta_1, \delta_2 \}$. Suppose $x \in S$, $\abs{x-c} \lt \delta$, and $x \not= c$. As $\delta > 0$ and $c$ is an cluster point, such an $x$ exists. Then
$\abs{L_1 - L_2}$ $= \abs{L_1 - f(x) + f(x) - L_2}\qquad \qquad\qquad\quad$
$\qquad \quad \leq \abs{L_1 - f(x)} + \abs{f(x) - L_2} $ $\ds\lt \frac{\vre}{2} + \frac{\vre}{2}$ $= \vre.$
Since $\abs{L_1-L_2} \lt \vre$ for arbitrary $\vre > 0$, then $L_1 = L_2.\;\bs$
4.3 Sequential limits
Theorem 4.3.1. Let $S \subset \R$, let $c$ be a cluster point of $S$, let $f \colon S \to \R$ be a function, and let $L \in \R$. Then $$f(x) \to L \;\text{ as }\; x \to c$$ if and only if for every sequence $\{ x_n \}$ of numbers such that $x_n \in S \setminus \{c\}$ for all $n$, and $\lim\, x_n = c$, we have that $$\{ f(x_n) \}\rightarrow L.$$
Example 4.3.1.
$\qquad\quad\; \displaystyle \lim_{x \to 0} \, \sin\left( \dfrac{1}{x} \right)$ $\qquad \;\;\;\;\displaystyle \lim_{x \to 0} \, x\sin\left( \dfrac{1}{x} \right)$
Example 4.3.1.
|
$\displaystyle \lim_{x\ra 0}f(x)$?
|
|---|
Example 4.3.1.
$f(x):=\sin (1/x)$
| Every $x_n\ra 0$ as $n\ra \infty$. | $f\left(x_n\right)$ converges or diverges? |
Example 4.3.1.
$f(x):=x\sin (1/x)$
| Every $x_n\to 0$ as $n\to \infty$. | $f\left(x_n\right)$ converges to $0$. |
Example 4.3.1.
Claim: $\;\displaystyle \lim_{x \to 0} \, \sin\left( \dfrac{1}{x} \right)$ does not exist, but $\displaystyle \lim_{x \to 0} \, x\sin\left( \dfrac{1}{x} \right)$ $ = 0.$
Proof.
For $\sin (1/x).$ Define a sequence
by
$x_n := \ds\frac{1}{\pi n + \dfrac{\pi}{2}}$
$\Rightarrow x_n=0$
$\;\Rightarrow \;\sin \left( \dfrac{1}{x_n} \right)
=
\sin \left(\pi n + \ds\frac{\pi}{2}\right)
$
$
= {(-1)}^n .$
Therefore, $\left\{ \sin \left( \dfrac{1}{x_n} \right) \right\}$ does not converge.
By Theorem 4.3.1,
$\lim_{x \to 0} \, \sin\left( \dfrac{1}{x} \right) $
does not exist. $\,\bs$
Example 4.3.1.
Claim: $\;\displaystyle \lim_{x \to 0} \, \sin\left( \dfrac{1}{x} \right)$ does not exist, but $\displaystyle \lim_{x \to 0} \, x\sin\left( \dfrac{1}{x} \right)$ $ = 0.$
Proof. Now for $x\sin (1/x).$ Let $\{ x_n \}$ be a sequence such that $x_n \not= 0$ for all $n$, and such that $\lim\, x_n = 0$. Note that $\abs{\sin(t)} \leq 1$ for all $t \in \R$. Then
$\abs{x_n\sin\left(\dfrac{1}{x_n}\right)-0}$ $= \abs{x_n}\abs{\sin\left(\dfrac{1}{x_n}\right)}$ $\leq \abs{x_n} .$
As $x_n \to 0$, then $\abs{x_n}\to 0$ $\;\Ra \left\{ x_n\sin\left(\dfrac{1}{x_n}\right) \right\}$ converges to zero.
By Theorem 4.3.1, $\;\ds \lim_{x \to 0} \, x\sin\left( \frac{1}{x} \right) = 0.\;\; \bs$
When a function does not have a limit?
Theorem 4.3.2 Let $S \subset \R$, let $c$ be a cluster point of $S$, let $f \colon S \to \R$ be a function, and let $L \in \R$.
if and only if
there exists a sequence $\{ x_n \}$ in $S$ with $x_n\neq c$ for all $n\in \N$ such that $\{ x_n \}$ converges to $c$ but the sequence $\{\, f(x_n) \}$ does not converge in $\R$.