课程： 2.8 复合功能的限度

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复合函数的界限
Composite functions are useful representations of how one quantity depends on a second quantity, and in turn that second quantity depends on a third quantity. The general representation of a composite function can be applied to lots of real-world analogies. But, what about the limit of a composite function? For example, your weekday commute time to work during a 4-hour time window in the morning may depend on the amount of traffic you encounter, which in turn depends on your departure time. Does it make sense to talk about what the limit of your commute time will be as the weekday departure time approaches a particular time in the 4-hour window? Does it make sense to talk about what the limit of your commute time will be in the limit as the departure time approaches a weekend time in the 4-hour window?
::复合函数是有用的表示, 表示一个数量取决于第二个数量, 而第二个数量则取决于第三个数量。复合函数的一般表示可以适用于许多真实世界的类比。但是, 复合函数的限度是什么? 比如, 您在早上四小时窗口工作的时间取决于您遇到的交通量, 而这反过来又取决于您离开的时间。谈到您在4小时窗口中一周出发的时间接近某个特定时间时段, 您的通勤时间在4小时窗口中接近一个特定时间时段的限度是什么是有道理的? 谈到您在4小时窗口中离开的时间接近一个周末时段, 您的通勤时间的限度是什么是有道理的?

Limits of Composite Functions
::复合函数的界限

Recall that the composition of functions $\begin{aligned} f (x) \end{aligned}$ and $\begin{aligned} g (x) \end{aligned}$ , indicated by the notation $\begin{aligned} (f \circ g) (x) \end{aligned}$ , means that the range of $\begin{aligned} g (x) \end{aligned}$ becomes the domain of $\begin{aligned} f (x) \end{aligned}$ . When the range of $\begin{aligned} g (x) \end{aligned}$ is in the domain of $\begin{aligned} f (x) \end{aligned}$ , there is no problem evaluating $\begin{aligned} (f \circ g) (x) \end{aligned}$ ; when that is not the case, $\begin{aligned} (f \circ g) (x) \end{aligned}$ is undefined.
::回顾函数f(x)和g(x)的构成,用符号(fg)(x)表示,这意味着 g(x) 的范围成为 f(x) 的域。当 g(x) 的范围在 f(x) 的域内时,没有问题评价 (fg)(x);如果不是这种情况,则没有定义 (fg(x)。

What does that mean for evaluating the limit of a composite function? We can illustrate with the following problems, two where the limit exists and one where the limit does not exist .
::这对评价复合函数的限度意味着什么?我们可以用下列问题来说明:两个问题存在限制,一个问题没有限制。

For $\begin{aligned} f (x) = \cos x \end{aligned}$ and $\begin{aligned} g (x) = 5 x^{2} \end{aligned}$ , find $\begin{aligned} lim_{x \to 0} (f \circ g) (x) \end{aligned}$ .
::f( x) =cosx 和 g( x) = 5x2, 找到 limx0( fg) (x) 。

We see that $\begin{aligned} (f \circ g) (x) = \cos (g (x)) = \cos (5 x^{2}) \end{aligned}$ . The limit can be evaluated by direct substitution:
:fg)(x)=cos(g(x))=cos(5x2)。

$\begin{aligned} lim_{x \to 0} (f \circ g) (x) = lim_{x \to 0} \cos (5 x^{2}) = \cos (0) = 1. \end{aligned}$

::立方公尺= 立方公尺= 立方尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方公尺

Note that the same result could have been obtained by evaluating:
::请注意,通过对以下各项进行评价,本可取得同样的结果:

$\begin{aligned} lim_{x \to 0} (f \circ g) (x) = lim_{x \to 0} \cos (5 x^{2}) = \cos (lim_{x \to 0} 5 x^{2}) = \cos (lim_{x \to 0} g (x)) = 1. \end{aligned}$

::立方公尺=立方公尺

In this example, the range of $\begin{aligned} g (x = 0) \end{aligned}$ is in the domain of $\begin{aligned} f (x) \end{aligned}$ .
::在此示例中, g(x=0) 的范围在 f(x) 的域内。

Now, consider $\begin{aligned} f (x) = \frac{1}{x + 1} \end{aligned}$ , $\begin{aligned} g (x) = x^{2} \end{aligned}$ . Find $\begin{aligned} lim_{x \to - 1} (f \circ g) (x) \end{aligned}$ .
::现在,考虑 f( x) = 1x+1, g( x) =x2. 查找 limx @% 1 (fg) (x) 。

We see that $\begin{aligned} (f \circ g) (x) = \frac{1}{x^{2} + 1} \end{aligned}$ and note that the quotient property of limits can be used because the denominator of the rational function is not zero. Hence by direct substitution we have $\begin{aligned} lim_{x \to - 1} (f \circ g) (x) = \frac{1}{(- 1)^{2} + 1} = \frac{1}{2} \end{aligned}$ .
::我们可以看到 (fg (x) = 1x2+1) 并且注意到, 限制的商数属性可以使用, 因为理性函数的分母不是零。因此, 通过直接替换, 我们有 limx1 (fg)(x) =1 (- 1) 2+1=12 。

Note that the same result could have been obtained by evaluating
::请注意,通过评价,本可取得同样的结果。

$\begin{aligned} lim_{x \to - 1} (f \circ g) (x) = lim_{x \to - 1} f (g (x)) = f (lim_{x \to - 1} g (x)) = f (1) = \frac{1}{2} \end{aligned}$

::limx1 (fg)(x) =limx1f (g(x)) = f(limx1g(x)) = f(1)=12

The above two problems illustrate the application of the following limit rule for composite functions:
::上述两个问题说明对复合职能适用下列限制规则:

$\begin{aligned} lim_{x \to a} (f \circ g) (x) = f (lim_{x \to a} g (x)) = f (b) \end{aligned}$

::limxa( fg)(x) = f (limxag(x)) = f(b)

Provided all of the following are true:
::只要以下所有情况属实:

$\begin{aligned} lim_{x \to a} g (x) = b \end{aligned}$
::limxag(x)=b

$\begin{aligned} f (b) \end{aligned}$ is defined
:b) 定义

$\begin{aligned} lim_{x \to b} f (x) \end{aligned}$ exists
::limxbf( x) 存在

$\begin{aligned} lim_{x \to b} f (x) = f (b) \end{aligned}$
::limxbf(x)=f(b)

The last three provisions (2-4) constitute the definition of “continuity” of the function $\begin{aligned} f (x) \end{aligned}$ at $\begin{aligned} x = b \end{aligned}$ . The fundamental concept of continuity will be explored later.
::最后三项规定(2-4)构成在x=b时f(x)职能的“连续性”的定义。以后将探讨连续性的基本概念。

Let’s see what happens when not all of the above provisions are met.
::让我们看看在未满足所有上述规定的情况下会发生什么。

Consider $\begin{aligned} f (x) = \frac{1}{x + 1} \end{aligned}$ , $\begin{aligned} g (x) = - 1 \end{aligned}$ . Find $\begin{aligned} lim_{x \to - 1} (f \circ g) (x) \end{aligned}$ .
::考虑 f(x) = 1x+1, g(x)\\\\\\\\\ 查找 limx\\ *1 (fg) (x) 。

Let’s see what happens when the above limit rule for composite functions is used:
::让我们看看在使用上述复合函数限制规则时会发生什么:

$\begin{aligned} lim_{x \to - 1} (f \circ g) (x) & = lim_{x \to - 1} f (g (x)) \\ = f (lim_{x \to - 1} g (x)) \\ = f (g (- 1)) \\ = f (- 1) \dots Undefined \\ = \frac{1}{0} \dots Undefined \end{aligned}$

::limx1 (fg)(x) =limx1f (g(x)) = f(limx1g(x)) = f(g(-1)) = f(-1) = f(-1). 未定义=10... 未定义

This result is undefined, and violates at least provision 2 of the limit rule for composite functions.
::这一结果没有界定,至少违反了综合职能限额规则第2条的规定。

Hence $\begin{aligned} lim_{x \to - 1} (f \circ g) (x) \end{aligned}$ does not exist.
::因此不存在limx1(fg)(x) 。

Examples
::实例

Example 1
::例1

Earlier, you were asked if it makes sense to talk about what the limit of your commute time will be as the weekday departure time approaches a particular time in the 4-hour window. Does it make sense to talk about what the limit of your commute time will be as the departure time approaches a weekend time in the 4-hour window?
::早些时候,有人问您是否应该谈论您通勤时间的限度是什么,因为每周的休假时间在4小时窗口中接近某一特定时间。谈到您通勤时间的限度是什么,4小时窗口中的周末时间接近一个周末,这是否合理?

In this example, commute time to work is dependent on amount of weekday traffic, which is in turn dependent on the time in a 4-hour window. It makes sense then, that the commute can be determined as weekday departure time approaches any time in the 4-hour window. It does not make sense, however, to talk about a commute time obtained by using a weekend time in the model. Weekend times, even if in the 4-hour window, are not in the domain of the commute time composite function.
::在这个例子中,上班通勤时间取决于每周交通量,这反过来又取决于4小时窗口中的时间。因此,在4小时窗口中,通勤时间可以随着每周休假时间的临近而确定。然而,在模型中使用周末时间谈论通勤时间是不合理的。周末时间,即使是在4小时窗口中,也不在通勤时间组合功能范围内。

Example 2
::例2

Find $\begin{aligned} lim_{x \to - \frac{1}{2}} e^{4 x + 3} \end{aligned}$ by identifying the components of this composite function and applying the limit of a composite function rule.
::通过确定该复合函数的构成部分,并适用复合函数规则的限度,查找limx12e4x+3。

The function $\begin{aligned} e^{4 x + 3} \end{aligned}$ can be partitioned into the composite components $\begin{aligned} f (x) = e^{x} \end{aligned}$ and $\begin{aligned} g (x) = 4 x + 3 \end{aligned}$ . The limit is then evaluated as follows:
::函数 e4x+3 可分割成复合组件 f(x)=ex 和 g(x)= 4x+3 。

$\begin{aligned} lim_{x \to - \frac{1}{2}} (f \circ g) (x) & = lim_{x \to - \frac{1}{2}} f (g (x)) \\ = e^{lim_{x \to - \frac{1}{2}} (4 x + 3)} \\ = e^{1} \\ lim_{x \to - \frac{1}{2}} e^{4 x + 3} & = e \approx 2.718 \end{aligned}$

::limx*%12 (fg)(x) =limx\\\12f(g(x)) = elimx= 12( 4x+3) = e1limx\\\\12e4x+3=e2. 718

Review
::回顾

For #1-5, identify two functions that make it a composite function.
::对于第1-5号,确定使它成为复合功能的两个功能。

$\begin{aligned} f (x) = x^{2} + 3 \end{aligned}$
:xx)=x2+3

$\begin{aligned} f (x) = \sqrt{x^{2} + 4 x} \end{aligned}$
:xx)=x2+4x

$\begin{aligned} f (x) = (x^{2} + 2 x + 10)^{\frac{3}{2}} \end{aligned}$
::f(x) = (x2+2x+10) 32

$\begin{aligned} f (x) = \sin (x^{2} + 3) \end{aligned}$
:xx) =sin(x2+3)

$\begin{aligned} f (x) = 2^{\sin x} \end{aligned}$
:xx)=2sinx

For #6-15, evaluate the limit.
::6 -15,评估限制。

$\begin{aligned} lim_{x \to 2} (x^{2} + 3) \end{aligned}$
::磅=2(x2+3)

$\begin{aligned} lim_{x \to 3} (\sqrt{x^{2} + 4 x}) \end{aligned}$
::limx%3(x2+4x)

$\begin{aligned} lim_{x \to - 1} (x^{2} + 2 x + 10)^{\frac{3}{2}} \end{aligned}$
::limx1(x2+2x+10)32

$\begin{aligned} lim_{x \to 1} (\sqrt{2 x^{3} + 3 x^{2} + 7}) \end{aligned}$
::limx%1( 2x3+3x2+7)

$\begin{aligned} lim_{x \to \sqrt{\frac{π}{2}}} \sin (x^{2} + π) \end{aligned}$
::立方公尺 [x2]

$\begin{aligned} lim_{x \to \frac{π}{2}} 2^{\sin x} \end{aligned}$
::22sinx

$\begin{aligned} lim_{x \to - 1} \sin (2^{x} \frac{π}{2}) \end{aligned}$
:2x% 2) 2

$\begin{aligned} lim_{x \to 3} (\sqrt[3]{2 x^{2} - 10}) \end{aligned}$
::立方厘米3( 2x2-2- 103)

$\begin{aligned} lim_{x \to - 1} e^{2 t^{2}} \end{aligned}$
::立方公尺#%1e2t2

$\begin{aligned} lim_{x \to - 1} \frac{x^{3} + 1}{x + 1} \end{aligned}$
::limx=%1x3+1x+1

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

常规

复合函数的界限

Limits of Composite Functions ::复合函数的界限

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Limits of Composite Functions
::复合函数的界限

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)