Honors Calculus Notes for [2 Functions]

Definition

Def. 2.1.1 Rigorous definition of limits

We say $\lim_{x\rightarrow a} f(x)=L$, where $a,L\in\mathbb R$, if the following holds:

$\forall \epsilon>0, \exists \delta >0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.

If there exists such an $L\in \mathbb R$ such that the above holds, we say $\lim_{x\rightarrow a}f(x) $ exists.

Def. 2.1.2 One-sided limits

We say $\lim_{x\rightarrow a^+} f(x)=L$, where $a,L\in\mathbb R$, if the following holds:

$\forall \epsilon>0, \exists \delta >0$ such that if $0<x-a<\delta$, then $|f(x)-L|<\epsilon$.

We say $\lim_{x\rightarrow a^-} f(x)=L$, where $a,L\in\mathbb R$, if the following holds:

$\forall \epsilon>0, \exists \delta >0$ such that if $-\delta<x-a<0$, then $|f(x)-L|<\epsilon$.

Def. 2.2 Infinity

We define $\lim_{x\rightarrow a}f(x)=+\infty$ if the following holds:

$\forall K>0, \exists \delta>0$ such that if $0<|x-a|<\delta$, then $f(x)>K$.

We define $\lim_{x\rightarrow a}f(x)=-\infty$ if the following holds:

$\forall K>0, \exists \delta>0$ such that if $0<|x-a|<\delta$, then $f(x)<-K$.

We define $\lim_{x\rightarrow +\infty}f(x)=L$ if the following holds:

$\forall \epsilon>0, \exists K>0$ such that if $x>K$, then $|f(x)-L|<\epsilon$.

We define $\lim_{x\rightarrow -\infty}f(x)=L$ if the following holds:

$\forall \epsilon>0, \exists K<0$ such that if $x<K$, then $|f(x)-L|<\epsilon$.

Def. 2.3 Continuity

A function $f$ is said to be continuous at $x=a$ if $\lim_{x\rightarrow a} f(x)$ exists, $f(a)$ is defined, and

\[\lim_{x\rightarrow a} f(x)=f(a) \]

If $f$ is continuous at every point $a$ in a set $S\subset \mathbb R$, then we say $f$ is continuous on $S$.

Def. 2.4 Domain, Codomain, Range

The domain of a function $f(x)$ means the set of allowable inputs.

If $D$ is the domain of a function $f(x)$, then any (non-empty) subset $E$ of $D$ can also taken to be the domain of $f(x)$. The domain is usually taken as the largest set in which $f(x)$ is defined.

The codomain of a function $f(x)$ is the set where the outputs belong to.

It does not mean that everything in the codomain is a possible output.

The range of a function $f:D\rightarrow C$ is the set of all achievable outputs. It is usually denoted by $f(D)$ where $D$ is the domain of $f$. In set notations, the range of $f$ is defined by:

\[f(D)=\lbrace f(x):x\in D \rbrace \]

Def. 2.5 Well-definedness

If $f(x)$ has the property that every input in the domain has exactly one output, then we say $f$ is well-defined, and $f$ qualifies to be a function.

Def. 2.6 Injective Function

A function $f:D\rightarrow C$ is said to be injective (or one-to-one) if

\[a\neq b \text{ in } D\Rightarrow f(a)\neq f(b). \]

Equivalently:

\[f(a)=f(b)\text{ where } a, b \text{ in } D\Rightarrow a=b \]

Def. 2.7 Increasing and decreasing functions.

Let $I$ be an interval in the domain $D$ of a function $f:D\rightarrow\mathbb R$.

A function $f$ is said to be strictly increasing on an interval $I$ if

\[f(x_1)<f(x_2) \text{ for any } x_1, x_2\in I \text{ such that } x_1<x_2 \]

A function $f$ is said to be strictly decreasing on an interval $I$ if

\[f(x_1)>f(x_2) \text{ for any } x_1, x_2\in I \text{ such that } x_1<x_2 \]

Def. 2.8 Surjective Functions

A function $f:D\rightarrow C$ is said to be surjective (or onto) if for any value $y$ in the codomain $C$, there is at least one $x$ in the domain $D$ such that $y=f(x)$.

Def. 2.9 Bijective Functions

A function $f:D\rightarrow C$ is said to be bijective if it is both injective and surjective.

Def. 2.9 Inverse Function

Let $f:D\rightarrow C$ be an injective function, then there exists an inverse function $f^{-1}:f(D)\rightarrow D$ which is defined by the rule:

\[f^{-1}(y)=x\Leftrightarrow y=f(x) \]

for any $y\in f(D).$

Def. 2.10 Identity Function

The identity function with domain $D$ is a function $id_D:D\rightarrow D$ defined by:

\[id_D(x)=x \]

for any $x$ in $D$.

Proposition & Theorem

Pro. 2.1

Let $L\in\mathbb R$. We have $\lim_{x\rightarrow a} f(x)=L$ if and only if $\lim_{x\rightarrow a^-} f(x)=L=\lim_{x\rightarrow a^+} f(x)$.

Proof Pro 2.1

$(\Rightarrow)$ is trivial. We only need to prove $(\Leftarrow)$.

$\forall \epsilon >0, \exists \delta_1>0$ such that $\forall 0<x-a<\delta_1, |f(x)-L|<\epsilon$. [Def 2.1.2]

$\forall \epsilon >0, \exists \delta_2>0$ such that $\forall -\delta_2<x-a<0, |f(x)-L|<\epsilon$. [Def 2.1.2]

Take $\delta=\min\lbrace \delta_1, \delta_2\rbrace>0, \forall \epsilon>0, \forall |x-a|<\delta, -\delta_2\leq-\delta<x-a<\delta\leq \delta_1, |f(x)-L|<\epsilon$.

Pro. 2.2 Arithmetic rules

Let $a\in\mathbb R$. Suppose $\lim_{x\rightarrow a}f(x)=L$ and $\lim_{x\rightarrow a} g(x)=M$ both exist ($L,M\in\mathbb R$), then the following holds.

(a) $\lim_{x\rightarrow a}(f(x)\pm g(x))=L\pm M$

(b) $\lim_{x\rightarrow a}(f(x)g(x))=LM$

These still holds if we replace $x\rightarrow a$ by $x\rightarrow a^+,x\rightarrow a^-,x\rightarrow +\infty,x\rightarrow -\infty$.

Proof Pro. 2.2(a)

\[|(a_n+b_n)-(L+M)|=|(a_n-L)+(b_n-M)|\leq |a_n-M|+|b_n-M|<\frac{\epsilon}{2}+\frac{\epsilon}{2} \]

Proof Pro. 2.2(b)

\[|a_nb_n-LM|=|(a_n-L)b_n+L(b_n-M)|\leq |a_n-L||b_n|+|L||b_n-M| \]

Pro. 2.3

If $\lim_{x\rightarrow a}g(x)=\pm\infty$, then $\lim_{x\rightarrow a}\frac{1}{g(x)}=0$.
If $\lim_{x\rightarrow a}g(x)=0$ and $g(x)>0$ when $x$ is sufficiently close to $a$, then $\lim_{x\rightarrow a}\frac{1}{g(x)}=+\infty$.
If $\lim_{x\rightarrow a}g(x)=0$ and $g(x)<0$ when $x$ is sufficiently close to $a$, then $\lim_{x\rightarrow a}\frac{1}{g(x)}=-\infty$.
If $\lim_{x\rightarrow a}f(x)=+\infty$ and $\lim_{x\rightarrow a}g(x)=+\infty$, then $\lim_{x\rightarrow a} (f(x)+g(x))=+\infty$.
If $\lim_{x\rightarrow a}f(x)=-\infty$ and $\lim_{x\rightarrow a}g(x)=-\infty$, then $\lim_{x\rightarrow a} (f(x)+g(x))=-\infty$.
If $\lim_{x\rightarrow a}f(x)=\infty$ and $\lim_{x\rightarrow a}g(x)=L\in \mathbb R$, then $\lim_{x\rightarrow a} (f(x)+g(x))=+\infty$.
If $\lim_{x\rightarrow a}f(x)=+\infty$ and $\lim_{x\rightarrow a}g(x)=L>0$, then $\lim_{x\rightarrow a} f(x)g(x)=+\infty$.
If $\lim_{x\rightarrow a}f(x)=+\infty$ and $\lim_{x\rightarrow a}g(x)=L<0$, then $\lim_{x\rightarrow a} f(x)g(x)=-\infty$.

These still holds if we replace $x\rightarrow a$ by $x\rightarrow a^+,x\rightarrow a^-,x\rightarrow +\infty,x\rightarrow -\infty$.

Pro. 2.4 Squeeze theorem

Suppose there exists $\delta>0$ such that $f(x)\leq g(x)\leq h(x)$ for any $x\in(a-\delta, a+\delta)$, and $\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}h(x)=L,$ then we have $\lim_{x\rightarrow a}g(x)=L$.

Proof Pro. 2.4

$\forall \epsilon>0, \exists\delta_1>0$ such that $\forall 0<|x-a|<\delta_1, |f(x)-L|<\epsilon$. [Def 2.1.1]

$\forall \epsilon>0, \exists\delta_2>0$ such that $\forall 0<|x-a|<\delta_2, |h(x)-L|<\epsilon$. [Def 2.1.1]

$\therefore \forall \epsilon>0, \exists\delta^{\prime}=\min\lbrace \delta, \delta_1, \delta_2\rbrace$ such that $\forall 0<|x-a|<\delta^{\prime}, L-\epsilon<f(x)-L<g(x)-L<h(x)-L<L+\epsilon$.

$\therefore\lim_{x\rightarrow a}g(x)=L.$

Pro. 2.5 Comparison rule

Suppose $\lim_{x\rightarrow a} f(x)$ and $\lim_{x\rightarrow a} g(x)$ exists, then the following implications hold:

(a) If $\lim_{x\rightarrow a} f(x)<\lim_{x\rightarrow a} g(x)$, then $\exists \delta>0$ such that $f(x)<g(x)$ whenever $0<|x-a|<\delta$.

(b) If $\forall \delta >0, \exists x_\delta\in(a-\delta, a+\delta)\setminus\lbrace a\rbrace$ such that $f(x_\delta)>g(x_\delta)$, then $\lim_{x\rightarrow a} f(x)\geq \lim_{x\rightarrow a} g(x)$.

Pro. 2.6

If $\lim_{x\rightarrow a}f(x)=+\infty$, $\lim_{x\rightarrow a}g(x)=L\in R$, and $h(x)$ is a function defined near $x=a$, then we have the following results:

(a) $\exists \delta>0$ such that $g(x)<f(x)$ whenever $0<|x-a|<\delta$.

(b) If $\exists \delta>0$ such that $f(x)<h(x)$ whenever $0<|x-a|<\delta$, then $\lim_{x\rightarrow a} h(x)=+\infty$.

Proof Pro. 2.6(a)

$\forall \epsilon>0, \exists \delta_2>0$ such that if $0<|x-a|<\delta_2, |g(x)-L|<\epsilon$, which means $L-\epsilon<g(x)<L+\epsilon.$ [Def. 2.1.1]

For $K=(\max\lbrace L+\epsilon, 0\rbrace+1)>0, \exists \delta_1>0$ such that if $0<|x-a|<\delta_1, f(x)>K>L+\epsilon.$ [Def. 2.2]

Then let $\delta=\min\lbrace\delta_1, \delta_2\rbrace, \forall 0<|x-a|<\delta, g(x)<L+\epsilon<f(x)$

Proof Pro. 2.6(b)

$\forall K>0, \exists \delta_0>0, \text{such that} \forall 0<|x-a|<\delta_0, f(x)>K$. [Def. 2.2]

Let $\theta=\min\lbrace\delta, \delta_2\rbrace$, we have $\forall K>0, \forall 0<|x-a|<\theta, K<f(x)<h(x)$.

Then $\lim_{x\rightarrow a}h(x)=+\infty.$ [Def. 2.2]

Pro. 2.7

Suppose $f$ is continuous at $x=b\in \mathbb R$, and $\lbrace a_n\rbrace $ is a sequence such that $\lim_{n\rightarrow\infty}a_n=b$, then

\[\lim_{n\rightarrow \infty}f(a_n)=f\left(\lim_{n\rightarrow\infty}a_n\right)=f(b) \]

Similarly, suppose $g(x)$ is a function (not necessarily continuous) such that $\lim_{x\rightarrow a}g(x)=b$, then

\[\lim_{x\rightarrow a}f(g(x))=f\left(\lim_{x\rightarrow a}g(x)\right)=f(b) \]

Proof Pro. 2.7

Sequence

Given any $\epsilon>0$, by the continuity of $f$ at $b$, there exists $\delta>0$ such that if $|y-b|<\delta$, then $|f(y)-f(b)|<\epsilon$. Here we intentionally use $y$ as the variable to avoid confusion.

Since we are also given $\lim_{n\rightarrow \infty}a_n=b$, $\exists N>0$ such that if $n>N$, then $|a_n-b|<\delta$. Hence, when putting $y=a_n$ for $n>N$, the condition $|y-b|<\delta$ is satisfied, so from the above we conclude that

\[|f(a_n)-f(b)|<\epsilon\text{ .} \]

This shows $\lim_{n\rightarrow \infty} f(a_n)=f(b)$.

Function

Since we are also given $\lim_{x\rightarrow a}g(x)=b$, $\exists \epsilon>0$ such that if $0<|x-a|<\epsilon$, then $|g(x)-b|<\delta$. Hence, when putting $y=g(x)$ for $0<|x-a|<\epsilon$, the condition $|y-b|<\delta$ is satisfied, so from the above we conclude that

\[|f(g(x))-f(b)|<\epsilon\text{ .} \]

This shows $\lim_{n\rightarrow \infty} f(g(x))=f(b)$.

Pro. 2.8 Composition rule: sequence(function)

Suppose $b(n):\mathbb N\rightarrow \mathbb R$ is a sequence such that $\lim_{n\rightarrow\infty}b(n)=L$, where $L\in \mathbb R$, and $n(x):D\subset\mathbb R\rightarrow\mathbb N$ is a function of $x$ such that $\lim_{x\rightarrow a}n(x)=+\infty$. Then, we have

\[\lim_{x\rightarrow a}b(n(x))=L\text{ .} \]

Proof Pro. 2.8

$\forall \epsilon>0$, from $\lim_{n\rightarrow \infty}b(n)=L$, $\exists N>0$ such that if $n>N$, then $|b(n)-L|<\epsilon$. With this $N$, from $lim_{x\rightarrow a}n(x)=+\infty$, $\exists\delta>0$ such that if $0<|x-a|<\delta$, then $n(x)>N$. With $n(x)>N$, we would have $|b(n(x))-L|<\epsilon$.

Overall, we proved that $\forall \epsilon>0$, $\exists\delta>0$ such that if $0<|x-a|<\delta$, then $|b(n(x))-L|<\epsilon$. It concludes that

\[\lim_{x\rightarrow a}b(n(x))=L\text{ .} \]

Pro. 2.9 Composition rules: function(function) and function(sequence)

Suppose $\lim_{y\rightarrow b} f(x)=L$ (here $f$ is not assumed to be continuous at $y=b$). Suppose $g(x)$ is a function such that

\[\lim_{x\rightarrow a} g(x)=b\text{ and } \exists \delta>0 \text{ such that } g(x)\neq b \text{ when } x\in (a-\delta, a+\delta)\setminus\lbrace a\rbrace \]

Then, we have

\[\lim_{x\rightarrow a}f(g(x))=L \]

Similarly, suppose $\lbrace a_n\rbrace$ is a sequence such that

\[\lim_{n\rightarrow \infty}a_n=b\text{ and } \exists N>0 \text{ such that } a_n\neq b \text{ whenever } n>N \]

Then, we have

\[\lim_{n\rightarrow \infty}f(a_n)=L \]

Proof Pro. 2.9

function(function)

From $\lim_{y\rightarrow b} f(y)=L$, $\forall \epsilon>0$, $\exists \delta>0$ such that if $0<|y-b|<\delta, |f(y)-L|<\epsilon$.

From $\lim_{x\rightarrow a} g(x)=b$, $\exists\delta_2>0$ such that $\forall 0<|x-a|<\delta_2$, $|g(x)-b|<\delta$.

$\therefore \exists\delta_2>0$ such that $\forall 0<|x-a|<\delta_2$, $|f(g(x))-L|<\epsilon$.

$\therefore \lim_{x\rightarrow a}f(g(a))=L$.

function(sequence)

From $\lim_{y\rightarrow b} f(y)=L$, $\forall \epsilon>0$, $\exists \delta>0$ such that if $0<|y-b|<\delta, |f(y)-L|<\epsilon$.

From $\lim_{n\rightarrow \infty} a_n=b$, $\exists N >0$ such that $\forall n>N$, $|a_n-b|<\delta$.

$\therefore \exists N>0$ such that $\forall n>N$, $|f(a_n)-L|<\epsilon$.

$\therefore \lim_{n\rightarrow \infty}f(a_n)=L$.

Pro. 2.10 Archimedean Principle

$\forall x, y>0, \exists n\in\mathbb {N} \text{ such that } nx>y.$

Proof Pro. 2.10

Assume on the contraty that $nx\leq y$ for all $n\in\mathbb N$, then the set

\[S:=\lbrace nx:n\in\mathbb N\rbrace \]

would be bounded from above by $y$. The Completeness Axiom shows $S$ has the least upper bound, denote by $\sup S\in\mathbb R$. Since it is the least upper bound, $\sup S -x$ is not an upper bound for $S$, so there exists $mx\in S$ (where $m\in\mathbb N$) such that $\sup S-x<mx$. However, it implies $\sup S<(m+1)x$. Since $(m+1)x\in S$, it contradicts to the fact that $\sup S$ is an upper bound for $S$.

Pro. 2.11 Density of rational numbers

For any $a, b\in \mathbb R$ such that $a<b$, there exists $r\in \mathbb Q$ such that $a<r<b$

Proof Pro. 2.11

Consider $b-a>0$, the Archimedean Principle asserts that there exists $n\in \mathbb N$ such that $n(b-a)>1$, or in other words, the open interval $(na, nb)$ has length $>1$. Then $(na, nb)$ mush contain an integer $m\in \mathbb Z$. It can be proved using the Well-Ordering Axiom of natural numbers. Then we have

\[na<m<nb\Rightarrow a<\frac{m}{n}<b \]

As $\frac{m}{n}\in Q$, we finish the proof.

Coro. 2.12

For any $x\in \mathbb R$, there exists a sequence $\lbrace r_n\rbrace$ of rational numbers such that $\lim_{n\rightarrow \infty}r_n=x$.

Proof Coro. 2.12

For each $n\in \mathbb N$, we consider the open interval $(x-\frac{1}{n}, x+\frac{1}{n})$. Density of rational numbers show $\exists r_n\in(x-\frac{1}{n}, x+\frac{1}{n})$. In other words, we have

\[x-\frac{1}{n}<r_n<x+\frac{1}{n}\text{ } \forall n\in \mathbb N \]

Since $x-\frac{1}{n}\rightarrow x$ and $x+\frac{1}{n}\rightarrow x$ as $n\rightarrow \infty$, the squeeze theorem shows $r_n\rightarrow x$ too.

Theo. 2.13 Intermediate value theorem

Suppose $f$ is continuous on $[a, b]$, and $f(a)<f(b)$. Suppose $L\in \mathbb R$ such that $f(a)<L<f(b)$, then $\exists c\in(a, b)$ such that $f(c)=L$.

Coro. 2.14

Suppose $f$ is continuous on $[a, b]$, and $f(a)f(b)<0$ (i.e. $f(a)$ and $f(b)$ have different sign), then $f$ has a root in $(a, b)$.

Pro. 2.15

Let $f:[a, b]\rightarrow \mathbb R$ be a continuous injective function on $[a, b]$, then $f$ is either strictly increasing or strictly decreasing on $[a, b]$.

Lemma 2.16

Suppose $f:[a, b]\rightarrow \mathbb R$ be a continuous injective function on $[a, b]$, and given $x_1, x_2, x_3\in[a, b]$ with $x_1<x_2<x_3$. Then, we have either $f(x_1)<f(x_2)<f(x_3)$ or $f(x_1)>f(x_2)>f(x_3)$

Pro. 2.17

Let $f:[a, b]\rightarrow \mathbb R$ be a continuous injective function on $[a, b]$, then $f^{-1}:f([a, b])\rightarrow [a, b]$ is continuous.

posted @ 2024-10-24 14:09 Displace 阅读(17) 评论(0) 收藏举报

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