### 数学代写|微积分代写Calculus代写|MAST10006

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|微积分代写Calculus代写|Limit of a Function at a Point

Definition 2.1.3 Let $f$ be a real valued function defined on a set $D \subseteq \mathbb{R}$, and let $a \in \mathbb{R}$ be a limit point of $D$. We say that $b \in \mathbb{R}$ is a limit of $f(x)$ as $x$ approaches $a$ or limit of $f$ at $a$, if for every $\varepsilon>0$, there exists $\delta>0$ such that
$$|f(x)-b|<\varepsilon \quad \text { whenever } x \in D, 0<|x-a|<\delta$$ Let us observe the following important property. Theorem 2.1.2 A function cannot have more than one limit at a given point. Proof Suppose $b_{1}$ and $b_{2}$ are limits of a function $f: D \rightarrow \mathbb{R}$ at a given point $a$, where $a$ is a limit point of $D$. Let $\varepsilon>0$ be given. By the definition of the limit, there exist $\delta_{1}>0$ and $\delta_{2}>0$ such that
\begin{aligned} &x \in D, 0<|x-a|<\delta_{1} \quad \Rightarrow \quad\left|f(x)-b_{1}\right|<\varepsilon \ &x \in D, 0<|x-a|<\delta_{2} \quad \Rightarrow \quad\left|f(x)-b_{2}\right|<\varepsilon \end{aligned}
Thus,
\begin{aligned} \left|b_{1}-b_{2}\right| &=\left|\left(b_{1}-f(x)\right)+\left(f(x)-b_{2}\right)\right| \ & \leq\left|b_{1}-f(x)\right|+\left|f(x)-b_{2}\right| \ &<2 \varepsilon \end{aligned}
whenever, $x \in D$, and $0<|x-a|<\delta:=\min \left{\delta_{1}, \delta_{2}\right}$
Notation 2.1.2 If $b$ is the limit of $f(x)$ as $x$ approaches $a$, the we denote this fact as
$$\lim _{x \rightarrow a} f(x)=b$$

## 数学代写|微积分代写Calculus代写|Limit of a Function in Terms of Sequences

Let $a$ be a limit point of $D \subseteq \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$. Suppose $\lim {x \rightarrow a} f(x)=b$. Since $a$ is a limit point of $D$, we know by Theorem $2.1 .1$ that there exists a sequence $\left(x{n}\right)$ in $D \backslash{a}$ such that $x_{n} \rightarrow a$. Does $f\left(x_{n}\right) \rightarrow b$ ? The answer is in the affirmative.
Theorem 2.1.6 If $\lim {x \rightarrow a} f(x)=b$, then for every sequence $\left(x{n}\right)$ in $D$ with $x_{n} \rightarrow a$, we have $f\left(x_{n}\right) \rightarrow b$.

Proof Suppose $\lim {x \rightarrow a} f(x)=b$. Let $\left(x{n}\right)$ be a sequence in $D$ such that $x_{n} \rightarrow a$. Let $\varepsilon>0$ be given. We have to show that there exists $N \in \mathbb{N}$ such that $\left|f\left(x_{n}\right)-b\right|<\varepsilon$ for all $n \geq N$.
Since $\lim {x \rightarrow a} f(x)=b$, we know that there exists $\delta>0$ such that $$x \in D, 0<|x-a|<\delta \Rightarrow|f(x)-b|<\varepsilon .$$ Also, since $x{n} \rightarrow a$, corresponding to the above $\delta$, there exists $N \in \mathbb{N}$ such that $\left|x_{n}-a\right|<\delta$ for all $n \geq N$. Hence, we have $\left|f\left(x_{n}\right)-b\right|<\varepsilon$ for all $n \geq N$.
The converse of the above theorem is also true.
Theorem 2.1.7 Suppose that, for every sequence $\left(x_{n}\right)$ in $D$ for which $x_{n} \rightarrow$, we have $f\left(x_{n}\right) \rightarrow b$. Then $\lim _{x \rightarrow a} f(x)=b$.

Proof Assume for a moment that $f$ does not have the limit $b$ as $x$ approaches $a$. Then, by Theorem 2.1.4, there exists $\varepsilon_{0}>0$ such that for every $\delta>0$, there exists at least one $x_{\delta} \in D$ such that
$$0<\left|x_{\delta}-a\right|<\delta \text { and }\left|f\left(x_{\delta}\right)-b\right| \geq \varepsilon_{0} .$$
In particular, for every $n \in \mathbb{N}$, there exists $x_{n} \in D$ such that
$$0<\left|x_{n}-a\right|<\frac{1}{n} \text { and }\left|f\left(x_{n}\right)-b\right| \geq \varepsilon_{0} .$$
Thus, $x_{n} \rightarrow a$ but $f\left(x_{n}\right) \nrightarrow b$. This is a contradiction to our hypothesis in the theorem.

Remark 2.1.2 Here are some implications of Theorem 2.1.6. Suppose $\left(x_{n}\right)$ is a sequence in $D \backslash{a}$ such that $x_{n} \rightarrow a$.

1. If $\left(f\left(x_{n}\right)\right)$ does not converge, then $\lim _{x \rightarrow a} f(x)$ does not exist.
2. If $\left(f\left(x_{n}\right)\right)$ does not converge to a given $b \in \mathbb{R}$, then either $\lim {x \rightarrow a} f(x)$ does not exist or $\lim {x \rightarrow a} f(x)$ exists but $\lim _{x \rightarrow a} f(x) \neq b$.
3. If $\left(y_{n}\right)$ is another sequence in $D \backslash{a}$ which converges to $a$ and the sequences $\left(f\left(x_{n}\right)\right)$ and $\left(f\left(y_{n}\right)\right)$ converge to different points, then $\lim _{x \rightarrow a} f(x)$ does not exist.

## 数学代写|微积分代写Calculus代写|Some Properties

For considering certain properties of the limit, and also for later use, we recall some definitions from set theory:

Suppose $f$ and $g$ are (real valued) functions with domains $D_{f}$ and $D_{g}$, respectively, and let $\alpha \in \mathbb{R}$. Suppose $D_{f} \cap D_{g} \neq \varnothing$. Then, we define functions $f+g, f g$ and $\alpha f$ as
\begin{aligned} (f+g)(x) &=f(x)+g(x), \quad x \in D_{f} \cap D_{g} \ (f g)(x) &=f(x) g(x), \quad x \in D_{f} \cap D_{g} \ (\alpha f)(x) &=\alpha f(x), \quad x \in D_{f} \end{aligned}
The function $f+g$ is called the sum of $f$ and $g$, and $f g$ is called the product of $f$ and $g$. If $f$ is a nonzero function, that is, $f(x) \neq 0$ for some $x \in D_{f}$, then we define the function $1 / f$ by
$$\left(\frac{1}{f}\right)(x)=\frac{1}{f(x)}, \quad x \in D_{f}^{\prime},$$
where $D_{f}^{\prime}:=\left{x \in D_{f}: f(x) \neq 0\right}$. Thus, we can also define the function $f / g$ by
$$\frac{f}{g}=f \frac{1}{g}$$
on the set $D_{f / g}:=\left{x \in D_{f} \cap D_{g}: g(x) \neq 0\right}$.
If $D:=\left{x \in D_{f}: f(x) \in D_{g}\right} \neq \varnothing$, then we define the composition of $g$ and $f$, denoted by $g \circ f$, by
$$(g \circ f)(x)=g(f(x)), \quad x \in D$$
Thus, domain of $g \circ f$ is the set $D_{g \circ f}:=\left{x \in D_{f}: f(x) \in D_{g}\right}$.
In the following, when we talk about the functions $f+g, f g, f / g$ and $g \circ f$, we mean their definitions as above with appropriate domains of definitions; we may not write the domains explicitly. Functions are assumed to be defined on subsets of the set $\mathbb{R}$ of real numbers, and are real valued.
The following three theorems can be proved using Theorems 2.1.6 and 2.1.7, and the results on convergence of sequences of real numbers (supply details). However, to get accustomed with the $\varepsilon-\delta$ arguments, we provide the detailed proof using the definition itself.

## 数学代写|微积分代写Calculus代写|Limit of a Function at a Point

|F(X)−b|<e 每当 X∈D,0<|X−一个|<d让我们观察以下重要性质。定理 2.1.2 一个函数在给定点不能有多个限制。证明假设b1和b2是函数的极限F:D→R在给定的点一个， 在哪里一个是一个极限点D. 让e>0被给予。根据极限的定义，存在d1>0和d2>0这样

X∈D,0<|X−一个|<d1⇒|F(X)−b1|<e X∈D,0<|X−一个|<d2⇒|F(X)−b2|<e

|b1−b2|=|(b1−F(X))+(F(X)−b2)| ≤|b1−F(X)|+|F(X)−b2| <2e

## 数学代写|微积分代写Calculus代写|Limit of a Function in Terms of Sequences

X∈D,0<|X−一个|<d⇒|F(X)−b|<e.另外，由于Xn→一个, 对应于上述d， 那里存在ñ∈ñ这样|Xn−一个|<d对所有人n≥ñ. 因此，我们有|F(Xn)−b|<e对所有人n≥ñ.

0<|Xd−一个|<d 和 |F(Xd)−b|≥e0.

0<|Xn−一个|<1n 和 |F(Xn)−b|≥e0.

1. 如果(F(Xn))不收敛，则林X→一个F(X)不存在。
2. 如果(F(Xn))不收敛到给定的b∈R，那么要么林X→一个F(X)不存在或林X→一个F(X)存在但林X→一个F(X)≠b.
3. 如果(是n)是另一个序列D∖一个收敛到一个和序列(F(Xn))和(F(是n))收敛到不同的点，然后林X→一个F(X)不存在。

## 数学代写|微积分代写Calculus代写|Some Properties

(F+G)(X)=F(X)+G(X),X∈DF∩DG (FG)(X)=F(X)G(X),X∈DF∩DG (一个F)(X)=一个F(X),X∈DF

(1F)(X)=1F(X),X∈DF′,

FG=F1G

(G∘F)(X)=G(F(X)),X∈D

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