# MATH 409 Lecture 9

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Lecture Slides

## Limit Supremum and Infimum

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

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

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

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

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

### Properties

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

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

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

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

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

## Limits of Functions

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

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

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

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

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

### Vs. Limits of Sequences

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

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

Proof. Suppose that ${\displaystyle f(x)\to L}$ as ${\displaystyle x\to a}$. Consider an arbitrary sequence ${\displaystyle \left\{x_{n}\right\}}$ of elements of the set ${\displaystyle I\setminus \left\{a\right\}}$ converging to ${\displaystyle a}$. For any ${\displaystyle \epsilon >0}$, there exists ${\displaystyle \delta >0}$ such that ${\displaystyle 0<\left|x-a\right|<\delta }$ implies ${\displaystyle \left|f(x)-L\right|<\epsilon }$ for all ${\displaystyle x\in \mathbb {R} }$. Further, there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle \left|x_{n}-a\right|<\delta }$ for all ${\displaystyle n\geq N}$. Then if ${\displaystyle \left|f(x_{n})-L\right|<\epsilon }$ for all ${\displaystyle n\geq N}$ [1]. Then ${\displaystyle \left|f(x_{n})-L\right|<\epsilon }$ for all ${\displaystyle n\geq N}$. We conclude that ${\displaystyle f(x_{n})\to L}$ as ${\displaystyle n\to \infty }$.

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

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

#### Comparison Theorem

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

#### Arithmetic Theorems

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

{\displaystyle {\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, ${\displaystyle M\neq 0}$, then

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

### Divergence to Infinity

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

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

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

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

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

### One-Sided Limits

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

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

${\displaystyle a

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

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

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

### Limits at Infinity

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

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

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

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

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

### Examples

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Footnotes

1. ${\displaystyle N}$ depends on ${\displaystyle \delta }$, which in turn depends on ${\displaystyle \epsilon }$.