Section: 替代查找限额 | 15.5 替代确定限额

Section outline

Introduction
::导言

Finding limits for the vast majority of points for a given function is as simple as substituting the number that $\begin{aligned} x \end{aligned}$ approaches into the function. Since this turns evaluating limits into an algebra-level substitution, most questions involving limits focus on the cases where substituting does not work.
::特定函数绝大多数点的查找限制与替换x进函数的数值一样简单,因为这将评价限制转换成代数级替代,大多数涉及限制的问题都集中在替换无效的情况下。

Suppose a painter working 100 feet above the ground drops his paint brush to the street. The velocity of the paint brush in feet per second at $\begin{aligned} t = 2 \end{aligned}$ is given by the following limit:
::假设一个在地面上工作100英尺的画家将油漆刷落到街上。

$\begin{aligned} v (t) = lim_{t \to 2} \frac{- 16 t^{2} + 64}{t - 2} . \end{aligned}$

::v(t) =limt2 - 16t2+64t-2。

Determine the velocity of the paint brush at $\begin{aligned} t = 2 \end{aligned}$ given by the limit analytically .
::以分析方式确定以 t=2 表示的画刷速度。

Using Substitution to Find Limits
::使用替代以查找限制

A 3rd approach to finding a limit is analytically using substitution. Finding a limit analytically means using algebraic approaches to find the limit.
::找到限制的第三种方法是在分析上使用替代方法,在分析上找到限制,在分析上找到使用代数法找到限制的方法。

If the function $\begin{aligned} f (x) \end{aligned}$ has no holes or an asymptote at $\begin{aligned} x = a \end{aligned}$ , then the limit of the function is equal to the function value.
::如果函数 f( x) 在 x=a 上没有空洞或小数,则函数的极限等于函数值。

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

::limxaf(x) = f(a)

Thus, you can substitute the value that $\begin{aligned} x \end{aligned}$ approaches into the function and evaluate the result. This approach works perfectly when there are no holes or asymptotes at that particular $\begin{aligned} x \end{aligned}$ -value, and you do not divide by zero when substituting .
::因此,您可以替换 x 进入函数的值, 并评价结果。当该 x 值没有空洞或零位时, 此方法会完全有效, 而替换时不会除以零。

However, occasionally there will be a hole or asymptote at $\begin{aligned} x = a \end{aligned}$ . If the function is a rational expression with a hole, then algebraically factor the numerator and denominator. Next, cancel any common factors in the numerator and denominator. Finally, substitute the value that $\begin{aligned} x \end{aligned}$ approaches into the resulting expression. Thus, t he limit in this case is the function value as if the hole did not exist.
::但是,在 x=a 时,偶尔会出现一个洞或零星。如果函数是带有洞的合理表达式,那么代数因素将乘以分子和分母。接下来,取消分子和分母中的任何共同系数。最后,将xaproache 的值替换为生成的表达式。因此,此情况下的限值是函数值,就像空洞不存在一样。

If no factors can be canceled or the function has an asymptote, the limit likely does not exist at that point. Try another approach to confirm this conclusion.
::如果无法取消任何因素, 或函数没有时点, 此时可能不存在限制。请尝试另一种方法来确认此结论。

An example of this approach can be seen in the following video:
::如下视频中可以看到这一方法的一个实例:

Play, Learn, and Explore to Determine Limits:
::游戏、学习和探索以确定限制 :

Examples
::实例

Example 1
::例1

Which of the limits below can you determine using direct substitution? Find that limit.
::使用直接替代,您能确定下限的哪个?

$\begin{aligned} lim_{x \to 2} \frac{x^{2} - 4}{x - 2}, lim_{x \to 3} \frac{x^{2} - 4}{x - 2} \end{aligned}$

::2x2 -4x-2, limx3x2 -4x-2, limx3x2 - 4x-2

Solution:
::解决方案 :

The limit on the right can be evaluated using direct substitution. The rational expression on the left has a hole at $\begin{aligned} x = 2 \end{aligned}$ , so it would need to be simplified first.
::右侧的限值可以通过直接替换来评估。左侧的理性表达式在 x=2 上有一个洞, 因此它需要先简化。

$\begin{aligned} lim_{x \to 3} \frac{x^{2} - 4}{x - 2} = \frac{3^{2} - 4}{3 - 2} = \frac{9 - 4}{1} = 5 \end{aligned}$

::3x2-4x-2=32-43-2=9-41=5

Example 2
::例2

Evaluate the following limit analytically:
::以分析方式评估以下范围:

$\begin{aligned} lim_{x \to 2} \frac{x^{2} - 4}{x - 2} . \end{aligned}$

::立方公尺2x2 -4x-2。

Solution:
::解决方案 :

$\begin{aligned} lim_{x \to 2} \frac{x^{2} - 4}{x - 2} & = lim_{x \to 2} \frac{(x - 2) (x + 2)}{(x - 2)} \\ = lim_{x \to 2} (x + 2) \\ = 2 + 2 \\ = 4 \end{aligned}$

::2x2x2-4x-2=limx2x2x-2(x+2)(x+2)(x-2)(x-2)=limx2x+2=2+2=2+2=4

Example 3
::例3

Evaluate the following limit analytically:
::以分析方式评估以下范围:

$\begin{aligned} lim_{x \to 4} \frac{x^{2} - x - 12}{x - 4} . \end{aligned}$

::4x2 -x -12x -4。

Solution:
::解决方案 :

$\begin{aligned} lim_{x \to 4} \frac{x^{2} - x - 12}{x - 4} & = lim_{x \to 4} \frac{(x - 4) (x + 3)}{(x - 4)} \\ = lim_{x \to 4} (x + 3) \\ = 4 + 3 \\ = 7 \end{aligned}$

::立方公尺xxx4x2-x-12x-4=limx4x4(x-4)(x+3)(x-4)=limx4(x+3)=4+3=7

Example 4
::例4

Recall the question from the Introduction: A painter working 100 feet above the ground drops his paint brush to the street. The velocity of the paint brush in feet per second at $\begin{aligned} t = 2 \end{aligned}$ is given by the following limit:
::回顾导言中的问题:在地上100英尺处工作的画家将油漆刷子扔到街上。

$\begin{aligned} v (t) = lim_{t \to 2} \frac{- 16 t^{2} + 64}{t - 2} . \end{aligned}$

::v(t) =limt2 - 16t2+64t-2。

Determine the velocity of the paint brush at $\begin{aligned} t = 2 \end{aligned}$ given by the limit analytically.
::以分析方式确定以 t=2 表示的画刷速度。

Solution:
::解决方案 :

$\begin{aligned} v (2) & = lim_{t \to 2} \frac{- 16 t^{2} + 64}{t - 2} \\ = lim_{t \to 2} \frac{- 16 (t^{2} - 4)}{t - 2} \\ = lim_{t \to 2} \frac{- 16 (t - 2) (t + 2)}{t - 2} \\ = lim_{t \to 2} - 16 (t + 2) \\ = - 64 ft/s \end{aligned}$

:2) = limt%2 - 16t2+64t-2=limt%2 - 16(t2- 4)t-2=limt%2 - 16(t-2)(t+2)(t+2)t-2-2=limt%2 - 16(t+2)_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Example 5
::例5

Evaluate the following limit analytically:
::以分析方式评估以下范围:

$\begin{aligned} lim_{x \to 3} \frac{x^{2} - 9}{x - 3} . \end{aligned}$

::立方厘米3x2-9x-3。

Solution:
::解决方案 :

$\begin{aligned} lim_{x \to 3} \frac{x^{2} - 9}{x - 3} & = lim_{x \to 3} \frac{(x - 3) (x + 3)}{(x - 3)} \\ = lim_{x \to 3} (x + 3) \\ = 6 \end{aligned}$

::立方厘米x3x2-9x-3=立方厘米x3-3(x-3)(x+3)(x-3)(x-3)=立方厘米x3(x+3)=6

Example 6
::例6

Evaluate the following limit analytically:
::以分析方式评估以下范围:

$\begin{aligned} lim_{t \to 4} \sqrt{t + 32} . \end{aligned}$

::limt4t+32.

Solution:
::解决方案 :

$\begin{aligned} lim_{t \to 4} \sqrt{t + 32} = \sqrt{4 + 32} = \sqrt{36} = 6 \end{aligned}$

::limt4t+32=4+32=36=6

Example 7
::例7

Evaluate the following limit analytically:
::以分析方式评估以下范围:

$\begin{aligned} lim_{y \to 4} \frac{3 | y - 1 |}{y + 4} . \end{aligned}$

::-43 yy -1 y+4。

Solution:
::解决方案 :

$\begin{aligned} lim_{y \to 4} \frac{3 | y - 1 |}{y + 4} = \frac{3 | 4 - 1 |}{4 + 4} = \frac{3 \cdot 3}{8} = \frac{9}{8} \end{aligned}$

::=34 -14+4=34 -14+4=338=98

Summary
::摘要

Substitution is a method of determining limits where the value that $\begin{aligned} x \end{aligned}$ is approaching is substituted into the function and the result is evaluated.
::替代是一种确定限度的方法,用以将x接近值替换为函数,并对结果进行评价。

If the function $\begin{aligned} f (x) \end{aligned}$ has no holes or asymptote at $\begin{aligned} x = a \end{aligned}$ , then $\begin{aligned} lim_{x \to a} f (x) = f (a) \end{aligned}$ .
::如果函数 f( x) 在 x=a 时没有孔或无音效, 然后是 limx+*af( x) = f( a) 。

If the function is a rational expression with a hole, then algebraically factor the numerator and denominator. Next, cancel any common factors in the numerator and denominator. Finally, substitute the value that $\begin{aligned} x \end{aligned}$ approaches into the resulting expression.
::如果函数是带有洞的合理表达式,则代数因素为分子和分母。接下来,取消分子和分母中的任何共同系数。最后,将x对结果表达式的数值替换为X。

If no factors can be canceled or the function has an asymptote, the limit likely does not exist at that point. Try another approach to confirm this conclusion.
::如果无法取消任何因素, 或函数没有时点, 此时可能不存在限制。请尝试另一种方法来确认此结论。

Review
::回顾

Evaluate the following limits analytically:
::分析评价以下限度:

1. $\begin{aligned} lim_{x \to 5} \frac{x^{2} - 25}{x - 5} \end{aligned}$
::1. limx=5x2-2-25x-5

2. $\begin{aligned} lim_{x \to - 1} \frac{x^{2} - 3 x - 4}{x + 1} \end{aligned}$
::2. limx%1x2-3x-4x+1

3. $\begin{aligned} lim_{x \to 5} \sqrt{5 x} - 12 \end{aligned}$
::3. limx55x-12

4. $\begin{aligned} lim_{x \to 0} \frac{x^{3} + 3 x^{2} - x}{5 x} \end{aligned}$
::4. limx=0x3+3x2-x5x

5. $\begin{aligned} lim_{x \to 1} \frac{3 x | x - 4 |}{x + 1} \end{aligned}$
::5. 立方公尺13xx-4x+1

6. $\begin{aligned} lim_{x \to 2} \frac{x^{2} + 5 x - 14}{x - 2} \end{aligned}$
::6. limx%2x2+5x-14x-2

7. $\begin{aligned} lim_{x \to 1} \frac{x^{2} - 8 x + 7}{x - 1} \end{aligned}$
::7. limx=1x2-8x+7x-1

8. $\begin{aligned} lim_{x \to 0} \frac{5 x - 1}{2 x^{2} + 3} \end{aligned}$
::8. limx05x-12x2+3

9. $\begin{aligned} lim_{x \to 1} 4 x^{2} - 2 x + 5 \end{aligned}$
::9. limx}14x2-2x+5

10. $\begin{aligned} lim_{x \to 0} \frac{x^{2} + 5 x}{x} \end{aligned}$
::10. limx=0x2+5xxx

11. $\begin{aligned} lim_{x \to - 3} \frac{x^{2} - 9}{x + 3} \end{aligned}$
::11. limx=%3x2-9x+3

12. $\begin{aligned} lim_{x \to 0} \frac{5 x + 1}{x} \end{aligned}$
::12. limx05x+1x

13. $\begin{aligned} lim_{x \to 1} \frac{5 x + 1}{x} \end{aligned}$
::13. limx=15x+1x

14. $\begin{aligned} lim_{x \to 5} \frac{x^{2} - 25}{x^{3} - 125} \end{aligned}$
::14. 立方×5x2-2-25x3-125

15. $\begin{aligned} lim_{x \to - 1} \frac{x - 2}{x + 1} \end{aligned}$
::15. limx%1x-2x+1

Review (Answers )
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

Introduction ::导言

Using Substitution to Find Limits ::使用替代以查找限制

Examples ::实例

lim x → 3 x 2 − 9 x − 3 = lim x → 3 ( x − 3 ) ( x + 3 ) ( x − 3 ) = lim x → 3 ( x + 3 ) = 6 ::立方厘米x3x2-9x-3=立方厘米x3-3(x-3)(x+3)(x-3)(x-3)=立方厘米x3(x+3)=6

Summary ::摘要

Review ::回顾

Review (Answers ) ::回顾(答复)

Introduction
::导言

Using Substitution to Find Limits
::使用替代以查找限制

Examples
::实例

$\begin{aligned} lim_{x \to 3} \frac{x^{2} - 9}{x - 3} & = lim_{x \to 3} \frac{(x - 3) (x + 3)}{(x - 3)} \\ = lim_{x \to 3} (x + 3) \\ = 6 \end{aligned}$

::立方厘米x3x2-9x-3=立方厘米x3-3(x-3)(x+3)(x-3)(x-3)=立方厘米x3(x+3)=6

Summary
::摘要

Review
::回顾

Review (Answers )
::回顾(答复)