## Calculus_微积分_Examples of limits at infinity

The evaluate-approximate-render (EAR) procedure is used for this type of limit as well.
Example 1 Determine $\lim x \rightarrow \infty \frac{2 x+8}{x^3+4 x+2}$. Solution We evaluate the expression at $x=\Omega$ :
$$\lim x \rightarrow \infty \frac{2 x+8}{x^3+4 x+2}=\frac{2 \Omega+8}{\Omega^3+4 \Omega+2} \quad \approx \frac{2 \Omega}{\Omega^3}=\frac{2}{\Omega^2}=2 \omega^2$$
The limit is 0 .
The graph of the function with the limit that is taken in example 1 has a horizontal asymptote of $y=0$ (on the right).
Example 2 Evaluate $\lim {x \rightarrow-\infty} \frac{x^2+4}{3 x^2+5 x-9}$ Solution We evaluate the expression at $x=-\Omega$ : $$\lim x \rightarrow-\infty \frac{x^2+4}{3 x^2+5 x-9}=\frac{(-\Omega)^2+4}{3(-\Omega)^2+5(-\Omega)-9}=\frac{\Omega^2+4}{3 \Omega^2-5 \Omega}$$ The limit is $\frac{1}{3}$. The graph of the function with the limit that is taken in example 2 has a horizontal asymptote of $y=\frac{1}{3}$ (on the left). Example 3 Find the limit: $\lim x \rightarrow \infty \frac{5 x^3-2 x+7}{x^2+7 x-1}$ Solution We evaluate the expression at $x=\Omega$ : $$\lim {x \rightarrow \infty} \frac{5 x^3-2 x+7}{x^2+7 x-1}=\frac{5 \Omega^3-2 \Omega+7}{\Omega^2+7 \Omega-1} \approx \frac{5 \Omega^3}{\Omega^2}=5 \Omega \doteq \infty$$
The limit is $\infty$.

## Calculus_微积分_Examples of finding asymptotes

Determining whether a function has a horizontal asymptote (or two) is accomplished by checking the limits at infinity. Since $\lim x \rightarrow \infty f(x)$ may be different from $\lim x \rightarrow-\infty f(x)$, both limits should be checked.
Example 5 Find all horizontal asymptotes on the graph of $f(x)=$ $\frac{3 x-7}{4 x+\sqrt[3]{x^3+5 x}}$
Solution We first check the right side by evaluating the limit as $x \rightarrow \infty$
$$\lim {x \rightarrow \infty} \frac{3 x-7}{4 x+\sqrt[3]{x^3+5 x}}=\frac{3 \Omega-7}{4 \Omega+\sqrt[3]{\Omega^3+5 \Omega}} \quad \approx \frac{3 \Omega}{4 \Omega+\sqrt[3]{\Omega^3}}=\frac{3}{4 \Omega}$$ The function has a horizontal asymptote on the right, $y=\frac{3}{5}$. We also need to check the left side to see if the result is different. We therefore evaluate the limit as $x \rightarrow-\infty$ : $$\lim {x \rightarrow-\infty} \frac{3 x-7}{4 x+\sqrt[3]{x^3+5 x}}=\frac{3(-\Omega)-7}{4(-\Omega)+\sqrt[3]{(-\Omega)^3+5(-\Omega)}} \quad=\frac{}{-4 \Omega}$$
The function has a horizontal asymptote of $y=\frac{3}{5}$ on the left side as well.

Reading Exercise 15 Find any horizontal asymptotes on the graph of $y=\frac{4 x-1}{x+3}$

# 微积分代考

## Calculus_微积分_Examples of limits at infinity

$$\lim x \rightarrow \infty \frac{2 x+8}{x^3+4 x+2}=\frac{2 \Omega+8}{\Omega^3+4 \Omega+2} \approx \frac{2 \Omega}{\Omega^3}=\frac{2}{\Omega^2}=2 \omega^2$$

$$\lim x \rightarrow-\infty \frac{x^2+4}{3 x^2+5 x-9}=\frac{(-\Omega)^2+4}{3(-\Omega)^2+5(-\Omega)-9}=\frac{\Omega^2+4}{3 \Omega^2-5 \Omega-}$$

$$\lim x \rightarrow \infty \frac{5 x^3-2 x+7}{x^2+7 x-1}=\frac{5 \Omega^3-2 \Omega+7}{\Omega^2+7 \Omega-1} \approx \frac{5 \Omega^3}{\Omega^2}=5 \Omega \doteq \infty$$

## Calculus_微积分_Examples of finding asymptotes

$$\lim x \rightarrow \infty \frac{3 x-7}{4 x+\sqrt[3]{x^3+5 x}}=\frac{3 \Omega-7}{4 \Omega+\sqrt[3]{\Omega^3+5 \Omega}} \approx \frac{3 \Omega}{4 \Omega+\sqrt[3]{\Omega^3}}$$

$$\lim x \rightarrow-\infty \frac{3 x-7}{4 x+\sqrt[3]{x^3+5 x}}=\frac{3(-\Omega)-7}{4(-\Omega)+\sqrt[3]{(-\Omega)^3+5(-\Omega)}}=$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: