# 数学代写|微积分代写Calculus代写|MATH1051

## 数学代写|微积分代写Calculus代写|How Can a Limit Fail to Exist

We have seen that a limit fails to exist at a number $a$ if the left- and right-hand limits are not equal (as in Example 4). The next two examples illustrate additional ways that a limit can fail to exist.
EXAMPLE 5 Investigate $\lim _{x \rightarrow 0} \sin \frac{\pi}{x}$.
SOLUTION Notice that the function $f(x)=\sin (\pi / x)$ is undefined at 0 . Evaluating the function for some small values of $x$, we get
$$\begin{array}{rrrl} f(1)=\sin \pi=0 & f\left(\frac{1}{2}\right) & =\sin 2 \pi=0 \ f\left(\frac{1}{3}\right) & =\sin 3 \pi=0 & f\left(\frac{1}{4}\right) & =\sin 4 \pi=0 \ f(0.1) & =\sin 10 \pi=0 & f(0.01) & =\sin 100 \pi=0 \end{array}$$
Similarly, $f(0.001)=f(0.0001)=0$. On the basis of this information we might be tempted to guess that the limit is 0 , but this time our guess is wrong. Note that although $f(1 / n)=\sin n \pi=0$ for any integer $n$, it is also true that $f(x)=1$ for infinitely many values of $x$ (such as $2 / 5$ or 2/101) that approach 0 . You can see this from the graph of $f$ shown in Figure 8 .

The dashed lines near the $y$-axis indicate that the values of $\sin (\pi / x)$ oscillate between 1 and $-1$ infinitely often as $x$ approaches 0 .

Since the values of $f(x)$ do not approach a fixed number as $x$ approaches 0 ,
$$\lim _{x \rightarrow 0} \sin \frac{\pi}{x} \text { does not exist }$$
Examples 3 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is easy to guess the wrong value if we use inappropriate values of $x$, but it is difficult to know when to stop calculating values. And, as the discussion after Example 1 shows, sometimes calculators and computers give the wrong values. In the next section, however, we will develop foolproof methods for calculating limits.

Another way a limit at a number $a$ can fail to exist is when the function values grow arbitrarily large (in absolute value) as $x$ approaches $a$.

## 数学代写|微积分代写Calculus代写|Calculating Limits Using the Limit Laws

In Section $2.2$ we used calculators and graphs to guess the values of limits, but we saw that such methods don’t always lead to the correct answers. In this section we use the following properties of limits, called the Limit Laws, to calculate limits.

These five laws can be stated verbally as follows:

1. The limit of a sum is the sum of the limits.
2. The limit of a difference is the difference of the limits.
3. The limit of a constant times a function is the constant times the limit of the function.
4. The limit of a product is the product of the limits.
5. The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0 ).
It is easy to believe that these properties are true. For instance, if $f(x)$ is close to $L$ and $g(x)$ is close to $M$, it is reasonable to conclude that $f(x)+g(x)$ is close to $L+M$. This gives us an intuitive basis for believing that Law 1 is true. In Section $2.4$ we give a precise definition of a limit and use it to prove this law. The proofs of the remaining laws are given in Appendix F.

EXAMPLE 1 Use the Limit Laws and the graphs of $f$ and $g$ in Figure 1 to evaluate the following limits, if they exist.
(a) $\lim {x \rightarrow-2}[f(x)+5 g(x)]$ (b) $\lim {x \rightarrow 1}[f(x) g(x)]$
(c) $\lim {x \rightarrow 2} \frac{f(x)}{g(x)}$ SOLUTION (a) From the graphs of $f$ and $g$ we see that $$\lim {x \rightarrow-2} f(x)=1 \quad \text { and } \quad \lim {x \rightarrow-2} g(x)=-1$$ Therefore we have $$\begin{array}{rlr} \lim {x \rightarrow-2}[f(x)+5 g(x)] & =\lim {x \rightarrow-2} f(x)+\lim {x \rightarrow-2}[5 g(x)] \quad \text { (by Limit Law 1) } \ & =\lim {x \rightarrow-2} f(x)+5 \lim {x \rightarrow-2} g(x) \quad \text { (by Limit Law 3) } \ & =1+5(-1)=-4 \end{array}$$

(b) We see that $\lim {x \rightarrow 1} f(x)=2$. But $\lim {x \rightarrow 1} g(x)$ does not exist because the left and right limits are different:
$$\lim {x \rightarrow 1^{-}} g(x)=-2 \quad \lim {x \rightarrow 1^{+}} g(x)=-1$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Calculating Limits Using the Limit Laws

1. 和的极限是极限的总和。
2. 差异的极限是极限的差异。
3. 常数乘以函数的极限是常数乘以函数的极限。
4. 乘积的极限是极限的乘积。
5. 商的极限是极限的商（前提是分母的极限不为 0 ）。
很容易相信这些属性是真实的。例如，如果 $f(x)$ 接近 $L$ 和 $g(x)$ 接近 $M$ ，可以合理 地得出结论 $f(x)+g(x)$ 接近 $L+M$. 这为我们提供了相信法则 1 为真的直觉基 础。在节 $2.4$ 我们给出一个极限的精确定义，并用它来证明这个定律。其余定律的 证明在附录 F 中给出。
示例 1 使用极限定律和图形 $f$ 和 $g$ 在图 1 中评估以下限制（如果存在）。
(一个) $\lim x \rightarrow-2[f(x)+5 g(x)]$ (乙) $\lim x \rightarrow 1[f(x) g(x)]$
(C) $\lim x \rightarrow 2 \frac{f(x)}{g(x)}$ 解决方案 (a) 从图表 $f$ 和 $g$ 我们看到
$$\lim x \rightarrow-2 f(x)=1 \quad \text { and } \quad \lim x \rightarrow-2 g(x)=-1$$
因此我们有
$$\lim x \rightarrow-2[f(x)+5 g(x)]=\lim x \rightarrow-2 f(x)+\lim x \rightarrow-2[5 g(x)] \quad \text { (by Limit }$$
(b) 我们看到 $\lim x \rightarrow 1 f(x)=2$. 但 $\lim x \rightarrow 1 g(x)$ 不存在，因为左右极限不同:
$$\lim x \rightarrow 1^{-} g(x)=-2 \quad \lim x \rightarrow 1^{+} g(x)=-1$$

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