# 物理代写|理论力学作业代写Theoretical Mechanics代考|Functions and Limits

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Functions and Limits

By the term function $f(x)$ one understands the unique attribution of a dependent variable $y$ from the co-domain $W$ to an independent variable $x$ from the domain of definition $D$ of the function $f(x)$ :
$$y=f(x) \quad ; \quad D \subset \mathbb{R} \stackrel{f}{\longrightarrow} W \subset \mathbb{R}$$
We ask ourselves how $f(x)$ changes with $x$. All elements of the sequence
$$\left{x_n\right}=x_1, x_2, x_3, \cdots, x_n, \cdots$$

shall be from the domain of definition of the function $f$. Then for each $x_n$ there exists a
$$y_n=f\left(x_n\right)$$
and therewith a ‘new’ sequence $\left{f\left(x_n\right)\right}$.
Definition $f(x)$ possesses at $x_0$ a limiting value $f_0$, if for each sequence $\left{x_n\right} \rightarrow x_0$ holds:
$$\lim {n \rightarrow \infty} f\left(x_n\right)=f_0$$ That is written as: $$\lim {x \rightarrow x_0} f(x)=f_0$$
Examples
1.
$$f(x)=\frac{x^3}{x^3+x-1} \quad ; \quad \lim _{x \rightarrow \infty} f(x)=?$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Trigonometric Functions

It can be assumed that the trigonometric functions are well-known from schoolmathematics. Therefore, only the most important relations shall be compiled in this subsection.

Figure 1.3 illustrates that the angle $\varphi$ can be expressed not only by angular degrees $^{\circ}$, but equally uniquely also via the arc of the circle $s$ :
$$s=s(\varphi) \quad: \quad s\left(360^{\circ}\right)=2 \pi r ; s\left(180^{\circ}\right)=\pi r ; s\left(90^{\circ}\right)=\frac{\pi}{2} r ; \ldots$$
One introduces the dimensionless quantity
$$\begin{gathered} \varphi=\frac{s}{r} \quad \text { ‘radian’ } \ \varphi\left(^{\circ}\right)=360(180,90,45,1) \longrightarrow 2 \pi\left(\pi, \frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{180}\right) \mathrm{rad} \end{gathered}$$

Trigonometric functions
In the right-angled triangle in Fig. 1.4 $a$ and $b$, adjacent and opposite to angle $\alpha$, respectively, are called the leg (side, cathetus) and $c$ the hypothenuse. With these terms one defines:
\begin{aligned} & \sin \alpha=\frac{b}{c} \ & \cos \alpha=\frac{a}{c} \ & \tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{b}{a} \ & \cot \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{1}{\tan \alpha}=\frac{a}{b} \end{aligned}

# 理论力学代写

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Functions and Limits

$$y=f(x) \quad ; \quad D \subset \mathbb{R} \stackrel{f}{\longrightarrow} W \subset \mathbb{R}$$

$$\backslash \text { left }\left{x_{-} n \backslash r i g h t\right}=x_{-} 1, x_{-} 2, x_{-} 3, \backslash c d o t s, x_{-} n, \backslash c d o t s$$

$$y_n=f\left(x_n\right)$$

$$\lim n \rightarrow \infty f\left(x_n\right)=f_0$$

$$\lim x \rightarrow x_0 f(x)=f_0$$

1.
$$f(x)=\frac{x^3}{x^3+x-1} \quad ; \quad \lim _{x \rightarrow \infty} f(x)=?$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Trigonometric Functions

$$s=s(\varphi) \quad: \quad s\left(360^{\circ}\right)=2 \pi r ; s\left(180^{\circ}\right)=\pi r ; s\left(90^{\circ}\right)=\frac{\pi}{2} r ; \ldots$$

$$\varphi=\frac{s}{r} \quad \text { ‘radian’ } \varphi\left({ }^{\circ}\right)=360(180,90,45,1) \longrightarrow 2 \pi\left(\pi, \frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{180}\right) \mathrm{rad}$$

$$\sin \alpha=\frac{b}{c} \quad \cos \alpha=\frac{a}{c} \tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{b}{a} \quad \cot \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{1}{\tan \alpha}=\frac{a}{b}$$

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