Course: 14.5 寻找限制的合理化

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查找限制的合理化

Some limits cannot be evaluated directly by substitution and no factors immediately cancel. In these situations there is another algebraic technique to try called rationalization. With rationalization, you make the numerator and the denominator of an expression rational by using the properties of conjugate pairs.

How do you evaluate the following limit using rationalization?
::如何利用合理化来评估以下限制?

$\begin{aligned} lim_{x \to 16} \frac{\sqrt{x} - 4}{x - 16} \end{aligned}$
::立方公尺16x-4x-16

Using Rationalization to Find Limits
::使用合理化来查找限制

Rationalization generally means to multiply a rational function by a clever form of one in order to eliminate radical symbols or imaginary numbers in the denominator. Rationalization is also a technique used to evaluate limits in order to avoid having a zero in the denominator when you substitute.
::合理化通常意味着将理性功能乘以一种聪明的形式,以消除分母中的激进符号或想象数字。合理化也是一种用来评估限度的技术,以避免在替换时分母为零。

To do this, you will use the properties of conjugates.
::要做到这一点,您将使用共产物的属性。

Conjugates can be used to simplify expressions with a radical in the denominator:
::共产体可用于简化分母中带有激进表达式的表达式:

$\begin{aligned} \frac{5}{1 + \sqrt{3}} = \frac{5}{(1 + \sqrt{3})} \cdot \frac{(1 - \sqrt{3})}{(1 - \sqrt{3})} = \frac{5 - 5 \sqrt{3}}{1 - 3} = \frac{5 - 5 \sqrt{3}}{- 2} \end{aligned}$

Conjugates can be used to simplify with $\begin{aligned} i \end{aligned}$ in the denominator:
::在分母i 中,可使用 conjudges 来简化 :

$\begin{aligned} \frac{4}{2 + 3 i} = \frac{4}{(2 + 3 i)} \cdot \frac{(2 - 3 i)}{(2 - 3 i)} = \frac{8 - 12 i}{4 + 9} = \frac{8 - 12 i}{13} \end{aligned}$
::42+3i=4(2+3i)(2-3i)(2-3i)(2-3i)(3i)=8-12i4+9=8-12i13)

Here, they can be used to transform an expression in a limit problem that does not immediately factor to one that does immediately factor.
::在这里,它们可以用来将一个表达方式转换成一个限制问题,而这个限制问题并不立即对立即产生因素的表达方式产生影响。

$\begin{aligned} lim_{x \to 16} \frac{(\sqrt{x} - 4)}{(x - 16)} \cdot \frac{(\sqrt{x} + 4)}{(\sqrt{x} + 4)} = lim_{x \to 16} \frac{(x - 16)}{(x - 16) (\sqrt{x} + 4)} \end{aligned}$
:x+4)(x+4(x+4)=limx16(x-16)(x-16)(x-16)(x+4))

Now you can cancel the common factors in the numerator and denominator and use substitution to finish evaluating the limit.
::现在,您可以取消分子和分母中的常见系数,并使用替代来完成对极限的评估。

The rationalizing technique works because when you algebraically manipulate the expression in the limit to an equivalent expression, the resulting limit will be the same. Sometimes you must do a variety of different algebraic manipulations in order avoid a zero in the denominator when using the substitution method.
::理顺技术之所以有效,是因为当您对等表达式限制内的表达式进行代数操控时,结果的极限将是一样的。有时您必须进行各种不同的代数操控,以避免在使用替代方法时分母为零。

Examples
::实例

Example 1
::例1

In order to evaluate the limit of the following rational expression, you need to multiply by a clever form of 1 so that when you substitute there is no longer a zero factor in the denominator.
::为了评估以下合理表达的限度,您需要以1的巧妙形式乘以1, 这样当您替换时, 分母中不再有零系数。

\begin{aligned} lim_{x \to 16} \frac{\sqrt{x} - 4}{x - 16} & = lim_{x \to 16} \frac{(\sqrt{x} - 4)}{(x - 16)} \cdot \frac{(\sqrt{x} + 4)}{(\sqrt{x} + 4)} \\ = lim_{x \to 16} \frac{(x - 16)}{(x - 16) (\sqrt{x} + 4)} \\ = lim_{x \to 16} \frac{1}{(\sqrt{x} + 4)} \\ = \frac{1}{4 + 4} \\ = \frac{1}{8} \end{aligned}

x+4(x+4)=14+4=18)=18 (x+4)=14+4=18

Example 2
::例2

Evaluate the following limit: $\begin{aligned} lim_{x \to 3} \frac{x^{2} - 9}{\sqrt{x} - \sqrt{3}} \end{aligned}$ .
::评估以下限值: limx%3x2- 9x-3。

$\begin{aligned} lim_{x \to 3} \frac{(x - 3) (x + 3)}{(\sqrt{x} - \sqrt{3})} \cdot \frac{(\sqrt{x} + \sqrt{3})}{(\sqrt{x} + \sqrt{3})} & = lim_{x \to 3} \frac{(x - 3) (x + 3) (\sqrt{x} + \sqrt{3})}{(x - 3)} \\ = lim_{x \to 3} (x + 3) (\sqrt{x} + \sqrt{3}) \\ = 6 \cdot 2 \sqrt{3} \\ = 12 \sqrt{3} \end{aligned}$

x+3(x3)(x+3)(x+3)(x+3)(x+3)(x+3)(x-3)(x+3)(x+3)(x+3)(x+3)(x-3)(x-3)(x-3)(x-3)(3))=limx3(x+3)(x+3)(x+3)(x+3)=6__23=123)

Example 3
::例3

Evaluate the following limit: $\begin{aligned} lim_{x \to 7} \frac{\sqrt{x + 2} - 3}{x - 7} \end{aligned}$ .
::评估以下限值: limx7x+2-3x-7。

\begin{aligned} lim_{x \to 7} \frac{\sqrt{x + 2} - 3}{x - 7} & = lim_{x \to 7} \frac{(\sqrt{x + 2} - 3)}{(x - 7)} \cdot \frac{(\sqrt{x + 2} + 3)}{(\sqrt{x + 2} + 3)} \\ = lim_{x \to 7} \frac{(x + 2 - 9)}{(x - 7) \cdot (\sqrt{x + 2} + 3)} \\ = lim_{x \to 7} \frac{(x - 7)}{(x - 7) \cdot (\sqrt{x + 2} + 3)} \\ = lim_{x \to 7} \frac{1}{(\sqrt{x + 2} + 3)} \\ = \frac{1}{\sqrt{7 + 2} + 3} \\ = \frac{1}{6} \end{aligned}

x+2+3)(x+2+3) = 17+2+3= 17+3= 16

Example 4
::例4

Evaluate the following limit: $\begin{aligned} lim_{x \to 0} \frac{(2 + x)^{- 1} - 2^{- 1}}{x} \end{aligned}$ .
::评估以下限值: limx0(2+x)-1-2-1x。

\begin{aligned} lim_{x \to 0} \frac{(2 + x)^{- 1} - 2^{- 1}}{x} & = lim_{x \to 0} \frac{\frac{1}{x + 2} - \frac{1}{2}}{x} \cdot \frac{(x + 2) \cdot 2}{(x + 2) \cdot 2} \\ = lim_{x \to 0} \frac{2 - (x + 2)}{2 x (x + 2)} \\ = lim_{x \to 0} \frac{- x}{2 x (x + 2)} \\ = lim_{x \to 0} \frac{- 1}{2 (x + 2)} \\ = - \frac{1}{2 (0 + 2)} \\ = - \frac{1}{4} \end{aligned}

::立方公尺xxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxx2+2-1-1xx=limxxxxxxxx01x+2-12xxxxxxxxxxxxx2xxxxxxxx=lxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Example 5
::例5

Evaluate the following limit: $\begin{aligned} lim_{x \to 0} (\frac{3}{x \sqrt{9 - x}} - \frac{1}{x}) \end{aligned}$
::评价以下限值: limx0( 3x9- x- 1x)

$\begin{aligned} lim_{x \to 0} (\frac{3}{x \sqrt{9 - x}} - \frac{1}{x}) & = lim_{x \to 0} (\frac{3}{x \sqrt{9 - x}} - \frac{\sqrt{9 - x}}{x \sqrt{9 - x}}) \\ = lim_{x \to 0} (\frac{3 - \sqrt{9 - x}}{x \sqrt{9 - x}}) \\ = lim_{x \to 0} (\frac{(3 - \sqrt{9 - x})}{x \sqrt{9 - x}} \cdot \frac{(3 + \sqrt{9 - x})}{(3 + \sqrt{9 - x})}) \\ = lim_{x \to 0} (\frac{9 - (9 - x)}{3 x \sqrt{9 - x} + x (9 - x)}) \\ = lim_{x \to 0} (\frac{x}{x (3 \sqrt{9 - x} + (9 - x))}) \\ = lim_{x \to 0} (\frac{1}{3 \sqrt{9 - x} + (9 - x)}) \\ = \frac{1}{3 \sqrt{9 - (0)} + (9 - (0))} \\ = \frac{1}{3 \sqrt{9} + (9)} \\ = \frac{1}{3 (3) + (9)} \\ = \frac{1}{18} \end{aligned}$

::3-9-xxx9-x-1x) = limxx9-x9-x9-x9-xxx) = limx0xx0-0(3-9-x9-xxx) = limx0(3-9xx) 3+9-x9-x(3+9-x)(3+9-x) = limx00-0(9-x) = limxx-x9-xx) = xxxx(39-x) +(9-x) = lix0-(9-x) +(9-x) = 139+(9-(0) = 69+13(3)+(9) =118=

Summary

Rationalization is a technique used to evaluate limits in order to avoid having a zero in the denominator when you substitute.
::合理化是一种用来评估限度的技术,以避免在替代时分母为零。

The process involves multiplying a rational function by a form of one (known as the conjugate) to eliminate radical symbols or imaginary numbers in the denominator.
::这一过程涉及将一种理性功能乘以一种形式(称为共产体),以消除分母中的激进符号或虚构数字。

Rationalization transforms an expression in a limit problem that does not immediately factor to one that does, allowing for substitution to evaluate the limit.
::合理化在限制问题的表达方式中转换了一种表达方式,而这种表达方式并没有立即考虑到这样做的表达方式,从而允许以替代方式来评价这一限制。

Review
::回顾

Evaluate the following limits:
::评价以下限度:

1. $\begin{aligned} lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} \end{aligned}$
::1. limx%9x-3x-9

2. $\begin{aligned} lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} \end{aligned}$
::2. limx4x-2x-4

3. $\begin{aligned} lim_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1} \end{aligned}$
::3. limx1x+3-2x-1

4. $\begin{aligned} lim_{x \to 0} \frac{\sqrt{x + 3} - \sqrt{3}}{x} \end{aligned}$
::4. limx=0x+3-3-3x

5. $\begin{aligned} lim_{x \to 4} \frac{\sqrt{3 x + 4} - x}{4 - x} \end{aligned}$
::5. limx=43x+4-x4-x4-x

6. $\begin{aligned} lim_{x \to 0} \frac{2 - \sqrt{x + 4}}{x} \end{aligned}$
::6. limx02-x+4x

7. $\begin{aligned} lim_{x \to 0} \frac{\sqrt{x + 7} - \sqrt{7}}{x} \end{aligned}$
::7. limx=0x+7-7-7x

8. $\begin{aligned} lim_{x \to 16} \frac{16 - x}{4 - \sqrt{x}} \end{aligned}$
::8. 石化1616-x4-x

9. $\begin{aligned} lim_{x \to 0} \frac{x^{2}}{\sqrt{x^{2} + 12} - \sqrt{12}} \end{aligned}$
::9. limx=0x2x2+12-12

10. $\begin{aligned} lim_{x \to 2} \frac{\sqrt{2 x + 5} - \sqrt{x + 7}}{x - 2} \end{aligned}$
::10. limx=22x+5-x+7x-2

11. $\begin{aligned} lim_{x \to 1} \frac{1 - \sqrt{x}}{1 - x} \end{aligned}$
::11. 石化11-1-1-x1-x

12. $\begin{aligned} lim_{x \to \frac{1}{9}} \frac{9 x - 1}{3 \sqrt{x} - 1} \end{aligned}$
::12. 立方公尺199x-13x-1

13. $\begin{aligned} lim_{x \to 4} \frac{4 x^{2} - 64}{2 \sqrt{x} - 4} \end{aligned}$
::13. 立方瓦 - 44x2 - 642x - 4

14. $\begin{aligned} lim_{x \to 9} \frac{9 x^{2} - 90 x + 81}{9 - 3 \sqrt{x}} \end{aligned}$
::14. limx99x2-90x+819-3x

15. When given a limit to evaluate, how do you know when to use the rationalization technique? What will the function look like?
::15. 当给评估限制时,你怎么知道何时使用合理化技术?该功能是什么样子?

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。

Section outline

General

查找限制的合理化

Using Rationalization to Find Limits ::使用合理化来查找限制

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)