## 数学代写|微积分代写Calculus代写|Limit example with trig

Example 4 Find $\lim {x \rightarrow 0} x \sin \frac{1}{x}$. Solution We evaluate the expression at $x=0+\alpha$, which is the same as $x=\alpha$ : $$\lim {x \rightarrow 0} x \sin \frac{1}{x}=\alpha \sin \frac{1}{\alpha} .$$
At this point, perhaps we are stuck, because we do not know how to evaluate trigonometric functions at hyperreal numbers. However, the transfer principle does tell us something about the trig functions. In section $1.1$, we learned that since $-1 \leq \sin x \leq 1$ for any real number $x$, the same is true for any hyperreal number $x$. Therefore, $-1 \leq \sin \frac{1}{\alpha} \leq 1$. Then, multiplying through that inequality by $\alpha$ gives either
$$-\alpha \leq \alpha \sin \frac{1}{\alpha} \leq \alpha$$
or
$$-\alpha \geq \alpha \sin \frac{1}{\alpha} \geq \alpha,$$
depending on whether $\alpha$ is positive or negative. In either case, $\alpha \sin \frac{1}{\alpha}$ is between $\alpha$ and $-\alpha$, so the expression must be infinitesimal, and therefore renders the real result zero. The entire calculation can be written as follows:
$$\lim {x \rightarrow 0} x \sin \frac{1}{x}=\underbrace{\alpha \cdot \underbrace{\sin \frac{1}{\alpha}}{\begin{array}{c} \text { between } \ 1 \text { and }-1 \end{array}}}_{\text {infinitesimal }} \doteq 0 .$$
The limit is 0 .

## 数学代写|微积分代写Calculus代写|Limit example with polynomial

There is nothing in the definition of limit that requires us to find limits only at places where we are dividing by zero. The procedure is the same, regardless of whether it is at a place where we are dividing by zero.
Example 6 Find $\lim _{x \rightarrow 2}\left(x^2+7 x-3\right)$.

\begin{aligned} \lim {x \rightarrow 2}\left(x^2+7 x-3\right) &=(2+\alpha)^2+7(2+\alpha)-3 \ &=4+4 \alpha+\alpha^2+14+7 \alpha-3=15+11 \alpha+\alpha^2 \ & \approx 15 . \end{aligned} The limit is 15 . In this example, the function is defined at $x=2$. In fact, $f(2)=2^2+$ $7(2)-3=15$, and the value of the function matches the limit. In this case there is no hole in the graph (figure 9$)$; the point $(2,15)$ is on the graph. In the last several examples, the function is not defined at $x=b$, so there is no point at $x=b$; but, this is not the casc with this example. This idea is discussed again later in this chapter, in section 1.6. Reading Exercise 17 Find $\lim {x \rightarrow 2}\left(x^2+1\right)$

The definition of limit requires that we render the same real result for every infinitesimal $\alpha$. In our previous examples, this has happened, but sometimes it does not happen.
Example 7 Find $\lim {x \rightarrow 3} \frac{|x-3|}{x-3}$ Solution We evaluate the expression at $x=3+\alpha$ : $$\lim {x \rightarrow 3} \frac{|x-3|}{x-3}=\frac{|3+\alpha-3|}{3+\alpha-3}=\frac{|\alpha|}{\alpha} .$$
At this point we are stuck! To evaluate $|\alpha|$, we need to know whether $\alpha$ is positive or negative, but because $\alpha$ is an arbitrary infinitesimal, it could be either. If $\alpha>0$, we continue with
$$\frac{|\bar{\alpha}|}{\alpha}=\frac{\alpha}{\alpha}=1 .$$
If $\alpha<0$, we continue with
$$\frac{|\alpha|}{\alpha}=\frac{-\alpha}{\alpha}=-1$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Limit example with trig

$$-\alpha \leq \alpha \sin \frac{1}{\alpha} \leq \alpha$$

$$-\alpha \geq \alpha \sin \frac{1}{\alpha} \geq \alpha,$$

$$\lim x \rightarrow 0 x \sin \frac{1}{x}=\underbrace{\alpha \cdot \underbrace{\sin \frac{1}{\alpha}} \text { between } 1 \text { and }-1}_{\text {infinitesimal }} \doteq 0$$

## 数学代写|微积分代写Calculus代写|Limit example with polynomial

$$\frac{|\bar{\alpha}|}{\alpha}=\frac{\alpha}{\alpha}=1 .$$

$$\frac{|\alpha|}{\alpha}=\frac{-\alpha}{\alpha}=-1$$

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