MATH 409 Lecture 9

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Lecture Slides

Review

• Limit Points

Limit Supremum and Infimum

Let $\left\{x_{n}\right\}$ be a bounded sequence of real numbers. For any $n\in \mathbb {N}$, let $E_{n}$ denote the set of numbers of the form $x_{k}$, where $k\geq n$. ($E_{n}$ starts from the $n$^th term in the sequence; note that $E_{n+1}$ is nested within $E_{n}$, and $E_{n}$ has at most one more element than $E_{n+1}$; Another $x_{n}$ could occur later in the sequence).

The set $E_{n}$ is bounded, hence $\sup {E_{n}}$ and $\inf {E_{n}}$ exist. Observe that the sequence $\left\{\sup {E_{n}}\right\}$ is decreasing, the sequence $\left\{\inf {E_{n}}\right\}$ is increasing (since $E_{1}$, $E_{2}$, … are nested sets), and both are bounded. Therefore both sequences are convergent.

The limit of $\left\{\sup {E_{n}}\right\}$ is called the limit supremum of the sequence $\left\{x_{n}\right\}$ and is denoted $\limsup _{n\to \infty }x_{n}$.

The limit of $\left\{\inf {E_{n}}\right\}$ is called the limit infimum of the sequence $\left\{x_{n}\right\}$ and is denoted $\liminf _{n\to \infty }x_{n}$.

If the sequence is not bounded above, then $\limsup _{n\to \infty }=+\infty$. If the sequence is not bounded below, then $\liminf _{n\to \infty }=-\infty$.

Properties

$\liminf _{n\to \infty }x_{n}\leq \limsup _{n\to \infty }x_{n}$

Note we have $\inf {E_{n}}\leq \sup {E_{n}}$ for all subsequences $E_{n}$, therefore $\liminf _{n\to \infty }\leq \limsup _{n\to \infty }E_{n}$ if they exist.

$\liminf _{n\to \infty }x_{n}$ and $\limsup _{n\to \infty }x_{n}$ are limit points of the sequence $\left\{x_{n}\right\}$.

All limit points of $\left\{x_{n}\right\}$ are contained in the interval $\left[\liminf _{n\to \infty }x_{n},\limsup _{n\to \infty }x_{n}\right]$

The sequence $\left\{x_{n}\right\}$ converges to a limit $a$ if and only if $\liminf _{n\to \infty }x_{n}=\limsup _{n\to \infty }x_{n}=a$.

Limits of Functions

Let $I\subset \mathbb {R}$ be an open interval and $a\in I$. Suppose $f:E\to \mathbb {R}$ is a function defined on a set $E\supset I\setminus \left\{a\right\}$.

We say that the function $f$ converges to a limit $L\in \mathbb {R}$ at the point $a$ if for every $\epsilon >0$ there exists $\delta =\delta (\epsilon )>0$ such that

$0<\left|x-a\right|<\delta \implies \left|f(x)-L\right|<\epsilon \,\!$

Notation: $L=\lim _{x\to a}f(x)$ or $f(x)\to L$ as $x\to a$.

The set $(a-\delta ,a)\cup (a,a+\delta )$ is called the punctured δ-neighborhood of $a$. Convergence to $L$ means that, given $\epsilon >0$, the image of this set under the map $f$ is contained in the ε-neighborhood $(L-\epsilon ,L+\epsilon )$ of $L$ provided that $\delta$ is small enough.

Vs. Limits of Sequences

Theorem. Let $I$ be an open interval containing a point $a\in \mathbb {R}$ and $f$ be a function defined on $I\setminus \left\{a\right\}$. Then $f(x)\to L$ as $x\to a$ if and only if for any sequence $\left\{x_{n}\right\}$ of elements of $I\setminus \left\{a\right\}$,

$\lim _{n\to \infty }x_{n}=a\implies \lim _{n\to \infty }f(x_{n})=L\,\!$

Proof. Suppose that $f(x)\to L$ as $x\to a$. Consider an arbitrary sequence $\left\{x_{n}\right\}$ of elements of the set $I\setminus \left\{a\right\}$ converging to $a$. For any $\epsilon >0$, there exists $\delta >0$ such that $0<\left|x-a\right|<\delta$ implies $\left|f(x)-L\right|<\epsilon$ for all $x\in \mathbb {R}$. Further, there exists $N\in \mathbb {N}$ such that $\left|x_{n}-a\right|<\delta$ for all $n\geq N$. Then if $\left|f(x_{n})-L\right|<\epsilon$ for all $n\geq N$ . Then $\left|f(x_{n})-L\right|<\epsilon$ for all $n\geq N$. We conclude that $f(x_{n})\to L$ as $n\to \infty$.

Conversely, suppose that $f(x)\not \to L$ as $x\to a$. Then there exists $\epsilon >0$ such that for any $\delta >0$, the image of the punctured neighborhood $(a-\delta ,a)\cup (a,a+\delta )$ of the point $a$ under the map $f$ is not contained in $(L-\epsilon ,L+\epsilon )$. In particular, for any $n\in \mathbb {N}$ there exists a point $x_{n}\in \left(a-{\frac {1}{n}}\right)\cup \left(a,a+{\frac {1}{n}}\right)$ such that $x_{n}\in I$ and $\left|f(x_{n})-L\right|\geq \epsilon$. We have that hte sequence $\left\{x_{n}\right\}$ converges to $a$ and $x_{n}\in I\setminus \left\{a\right\}$. However, $f(x_{n})\not \to L$ as $n\to \infty$.

quod erat demonstrandum

Using this sequential characterization of limits, we can derive limit theorems for convergence of functions from analogous theorems dealing with convergence of sequences.

Limit Theorems

Squeeze Theorem

If $\lim _{x\to a}f(x)=\lim _{x\to a}g(x)=L$ and $f(x)\leq h(x)\leq g(x)$ for all $x$ in a punctured neighborhood of the point $a$, then $\lim _{x\to a}h(x)=L$.

Comparison Theorem

If $\lim _{x\to a}f(x)=L$ and $\lim _{x\to a}g(x)=M$

Arithmetic Theorems

If $\lim _{x\to a}f(x)=L$ and $\lim _{x\to a}g(x)=M$, then

${\begin{aligned}\lim _{x\to a}\left(f+g\right)\left(x\right)&=L+M\\\lim _{x\to a}\left(f-g\right)\left(x\right)&=L-M\\\lim _{x\to a}\left(f\,g\right)\left(x\right)&=L\,M\end{aligned}}$

If, additionally, $M\neq 0$, then

$\lim _{x\to a}\left({\frac {f}{g}}\right)\left(x\right)={\frac {L}{M}}$

Divergence to Infinity

Let $I\subset \mathbb {R}$ be an open interval and $a\in I$. Suppose $f:E\to \mathbb {R}$ is a function defined on the set $E\supset I\setminus \left\{a\right\}$.

We say that the function $f$ diverges to $+\infty$ at the point $a$ if for every $C\in \mathbb {R}$ there exists $\delta =\delta (C)>0$ such that

$0<\left|x-a\right|<\delta \implies f(x)>C$

Notation: $\lim _{x\to a}f(x)=+\infty$ or $f(x)\to +\infty$ as $x\to a$.

Similarly, divergence to $-\infty$ at the point $a$

One-Sided Limits

Let $f:E\to \mathbb {R}$ be a function defined on a set $E\subset \mathbb {R}$.

We say that $f$ converges to a right-hand limit $L\in \mathbb {R}$ at a point $a\in \mathbb {R}$ if the domain $E$ contains an interval $(a,b)$ and for every $\epsilon >0$ there exists $\delta =\delta (\epsilon )>0$ such that

$a<x<a+\delta \implies \left|f(x)-L\right|<\epsilon \,\!$

Notation: $L=\lim _{x\to a+}f(x)$.

Similarly, we define the left-hand-limit $\lim _{x\to a-}f(x)$.

Theorem. $f(x)\to L$ as $x\to a$ if and only if $\lim _{x\to a+}f(x)=\lim _{x\to a-}f(x)=L$.

Limits at Infinity

Let $f:E\to \mathbb {R}$ be a function defined on a set $E\subset \mathbb {R}$.

We say that $f$ converges to a limit $L\in \mathbb {R}$ as $x\to +\infty$ if the domain $E$ contains an interval $(a,+\infty )$ and for every $\epsilon >0$, there exists $C=C(\epsilon )\in \mathbb {R}$ such that

$x>C\implies \left|f(x)-L\right|<\epsilon \,\!$

Notation: $L=\lim _{x\to +\infty }f(x)$ or $f(x)\to L$ as $x\to +\infty$

Similarly we define the limit $\lim _{x\to -\infty }f(x)$.

Examples

Constant function. $f(x)=c$ for all $x\in \mathbb {R}$ and some $c\in \mathbb {R}$.

$\lim _{x\to a}f(x)=c$ for all $a\in \mathbb {R}$. Also, $\lim _{x\to \pm \infty }f(x)=c$.

Identity function. $f(x)=x$ for $x\in \mathbb {R}$.

$\lim _{x\to a}f(x)=a$ for all $a\in \mathbb {R}$. Also $\lim _{x\to \pm \infty }f(x)=\pm \infty$.

Heaviside function. Prof. defines it as $f(x)={\begin{cases}1&x>0\\0&x\leq 0\end{cases}}$

$\lim _{x\to 0+}=0$, and $\lim _{x\to 0-}=1$.

Harmonic function. $f:\mathbb {R} \setminus \left\{0\right\}\to \mathbb {R}$, where $f(x)={\frac {1}{x}}$

• $\lim _{x\to a}f(x)={\frac {1}{a}}$ for all $a\neq 0$.
• $\lim _{x\to 0+}=+\infty$
• $\lim _{x\to 0-}=-\infty$
• $\lim _{x\to \pm \infty }f(x)=0$

Sine. $f:\mathbb {R} \setminus \left\{0\right\}\to \mathbb {R}$, where $f(x)=\sin {\frac {1}{x}}$

$\lim _{x\to 0+}f(x)$ does not exist since $f((0,\delta ))=[-1,1]$ for any $\delta >0$.

$f:\mathbb {R} \setminus \left\{0\right\}\to \mathbb {R}$, where $f(x)=x\,\sin {\frac {1}{x}}$.

$\lim _{x\to 0}=0$ by squeeze theorem between $f(x)=x$ and $f(x)=-x$.

Dirichlet function. $f(x)={\begin{cases}1&x\in \mathbb {Q} \\0&x\in \mathbb {R} \setminus \mathbb {Q} \end{cases}}$

$\lim _{x\to a}f(x)$ does not exist since $f((c,d))=\left\{0,1\right\}$ for any interval $(c,d)$. In other words, both rational and irrational points are dense in $\mathbb {R}$.

Riemann function. $f(x)={\begin{cases}{\frac {1}{q}}&x={\frac {p}{q}}\ {\mbox{a reduced fraction}}\\0&x\in \mathbb {R} \setminus \mathbb {Q} \end{cases}}$.

$\lim _{x\to a}f(x)=0$ for all $a\in \mathbb {R}$ indeed, for any $n\in \mathbb {N}$ and a bounded interval $(c,d)$, there are only finitely many points $x\in (c,d)$ such that $f(x)\geq {\frac {1}{n}}$.

On the other hand, $\lim _{x\to +\infty }f(x)$ and $\lim _{x\to -\infty }f(x)$ do not exist.