15.6 寻找限制的合理化

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  查找限制的合理化

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  Introduction
  ::导言

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  Suppose you and a friend are  driving across the country on a road trip. The amount of time the two of you can drive is modeled with the following limit, where 16 hours is the most amount of time you and your friend can drive: 
  ::假设你和一位朋友在路上开车穿越国家。你们两人可以驾驶的驾驶时间有以下限制,其中16小时是你们和朋友可以驾驶的最长时间:

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  $$\begin{align*} \lim \limits_{x \to 16} \frac{\sqrt{x}-4}{x-16}.\end{align*}$$
  ::16x16x4x16。

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  How do you evaluate the limit using rationalization?
  ::你如何利用合理化来评估限额?

  

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  Rationalization to Find Limits
  ::查找限制的合理化

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  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.
  ::有些限制不能直接通过替代来评估,没有因素会立即取消。在这些情况下,还有另一种代数技术可以尝试合理化。有了合理化,你就可以使用同性配对的特性来使数字和表达的分母合理化。

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  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:

  $$\begin{align*}\frac{5}{1+ \sqrt{3}}=\frac{5}{\left ( 1+ \sqrt{3} \right )} \cdot \frac{\left ( 1- \sqrt{3} \right )}{\left ( 1- \sqrt{3} \right )}=\frac{5-5 \sqrt{3}}{1-3}=\frac{5-5 \sqrt{3}}{-2}.\end{align*}$$

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  Conjugates can be used to simplify complex numbers with $\begin{align*}i\end{align*}$ in the denominator:
  ::共产体可用于简化复杂数字,分母为i:

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

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  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,它们可用于将表达方式转换成一个限制问题,该限制问题不能立即对立即考虑的问题加以考虑:

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  $$\begin{align*}\lim \limits_{x \to 16} \frac{\left ( \sqrt{x}-4 \right )}{(x-16)} \cdot \frac{\left ( \sqrt{x}+4 \right )}{\left ( \sqrt{x}+4 \right )}=\lim \limits_{x \to 16} \frac{(x-16)}{(x-16)\left ( \sqrt{x}+4 \right )}.\end{align*}$$
  :xx+4)(x+4)(xx+4)=limx16(x-16)(x-16)(x-16)(x+4)。

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  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因子, 并使用替代来完成对极限的评估 。

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  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 to 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,您必须进行不同的代数操控,以避免在使用替代方法时分母为零。

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  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:

    

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  Examples
  ::实例

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  Example 1
  ::例1

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  Evaluate the following limit:
  ::评估以下限度:

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  $$\begin{align*}\lim_{x \to 3}\frac{x^2-9}{\sqrt{x}-\sqrt{3}}.\end{align*}$$
  ::立方公尺3x2 -9xx3

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  Solution:
  ::解决方案 :

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  lim x → 3 ( x − 3 ) ( x + 3 ) ( √ x − √ 3 ) ⋅ ( √ x + √ 3 ) ( √ x + √ 3 ) = lim x → 3 ( x − 3 ) ( x + 3 ) ( √ x + √ 3 ) ( x − 3 ) = lim x → 3 ( x + 3 ) ( √ x + √ 3 ) = 6 ⋅ 2 √ 3 = 12 √ 3

  $$\begin{align*}\lim_{x \to 3}\frac{(x-3)(x+3)}{\left(\sqrt{x}-\sqrt{3}\right)} \cdot \frac{\left(\sqrt{x}+\sqrt{3}\right)}{\left(\sqrt{x}+\sqrt{3}\right)}&=\lim_{x \to 3}\frac{(x-3)(x+3)\left(\sqrt{x}+\sqrt{3}\right)}{(x-3)} \\ &=\lim_{x \to 3}(x+3)\left(\sqrt{x}+\sqrt{3}\right) \\ &=6 \cdot 2\sqrt{3} \\ &=12 \sqrt{3}\end{align*}$$
  ::3x3(x-3)(x+3)(x+3)(xxx__3)(xx__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__x__3)(x__3)(x__3)(x__3)(x__3)(x__3)(x__3)(x__3)(x__3)(x__3)(x__3)(x__3)=6__2__3=12__3)。

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  Example 2
  ::例2

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  Evaluate the following limit: 

  $$\begin{align*}\lim_{x \to 25} \frac{x-25}{\sqrt{x}-5}.\end{align*}$$
  ::评估以下限额:limx25x-25x-5。

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  Solution:  
  ::解决方案 :

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  $$\begin{align*}\lim \limits_{x \to 25}\frac{x-25}{\sqrt{x}-5}&=\lim \limits_{x \to 25}\frac{(x-25)}{\left(\sqrt{x}-5\right)} \cdot \frac{\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)} \\ &=\lim \limits_{x \to 25} \frac{(x-25)\left(\sqrt{x}+5\right)}{(x-25)} \\ &=\lim \limits_{x \to 25}\left(\sqrt{x}+5\right) \\ &=\sqrt{25}+5 \\ &=10\end{align*}$$
  ::==10xx+5=10 =10 =10 =5xx+5=10 =10 =5x+5=5x=5x=5x=5x=25x=25x25(x+5=5x5=5x5x=25x25x+5=25x+5=25x+5=25x+5=25+5=5x=10

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  Example 3
  ::例3

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  Evaluate the following limit: 

  $$\begin{align*}\lim_{x \to 7}\frac{\sqrt{x+2}-3}{x-7}.\end{align*}$$
  ::评价以下限值: limx7x+2-3x-7。

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  Solution:
  ::解决方案 :

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  $$\begin{align*}\lim_{x \to 7}\frac{\sqrt{x+2}-3}{x-7}&=\lim_{x \to 7}\frac{\left(\sqrt{x+2}-3\right)}{(x-7)} \cdot \frac{\left(\sqrt{x+2}+3\right)}{\left(\sqrt{x+2}+3\right)} \\ &=\lim_{x \to 7}\frac{(x+2-9)}{(x-7) \cdot \left(\sqrt{x+2}+3\right)} \\ &=\lim_{x \to 7}\frac{(x-7)}{(x-7) \cdot \left(\sqrt{x+2}+3\right)} \\ &=\lim_{x \to 7}\frac{1}{\left(\sqrt{x+2}+3\right)} \\ &=\frac{1}{\sqrt{7+2}+3} \\ &=\frac{1}{6}\end{align*}$$
  :xx+2+2+3+3)(x-7)(x-7)(xxx+2+3)(xx+2+3)(xx+2+3)=limx7(x+2-9)(x-7)(x-7)(x-7)(x+2+3)=limx7(x-7)(x-7)(x-7)(x+2+3)=7(x-7)(x+2+3)=(x-7)(7)(x+2+3)=limx7(x-7)(x+2+3)==x7(x+2+3)=16

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  Example 4
  ::例4

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  Recall the question from the I ntroduction: The  amount of time you and your friend can drive is modeled with the following limit, where 16 hours is the most amount of time you and your friend can drive: 
  ::回顾导言中的问题:你和你的朋友可以驾驶的驾驶时间有以下限制,其中16小时是你和你的朋友可以驾驶的最长时间:

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  $$\begin{align*} \lim \limits_{x \to 16} \frac{\sqrt{x}-4}{x-16}\end{align*}$$
  ::立方公尺16x-4x-16

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  How do you evaluate the limit using rationalization?
  ::你如何利用合理化来评估限额?

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  Solution:
  ::解决方案 :

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  $$\begin{align*}\lim_{x \to 16}\frac{\sqrt{x}-4}{x-16}&=\lim_{x \to 16}\frac{\left(\sqrt{x}-4\right)}{(x-16)} \cdot \frac{\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)} \\ &=\lim_{x \to 16}\frac{(x-16)}{(x-16)\left(\sqrt{x}+4\right)} \\ &=\lim_{x \to 16}\frac{1}{\sqrt{x}+4} \\ &=\frac{1}{4+4} \\ &=\frac{1}{8}\end{align*}$$
  :x+4)(x+4)(x+4)(x+4)(xx-16)(x-16)(x-16)(x-16)(x+4)(x-16)(x-16)(x+4)(limx)=166(16)=161(16)x+1(x+4)+4=14+4=18)

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  Example 5
  ::例5

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  Evaluate the following limit: 

  $$\begin{align*}\lim_{x \to 0} \frac{(2+x)^{-1}-2^{-1}}{x}.\end{align*}$$
  ::评估以下限值: limx0(2+x)-1-2-1x。

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  Solution:
  ::解决方案 :

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  Even though the given limit does not have radical expressions, the rationalization process learned in this concept is similar to the process used to find  the given limit. 
  ::尽管给定的限制没有激进的表达方式,但这一概念中所学到的合理化进程与用来确定给定限制的程序相似。

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  lim x → 0 ( 2 + x ) − 1 − 2 − 1 x = lim x → 0 1 x + 2 − 1 2 x ⋅ ( x + 2 ) ⋅ 2 ( x + 2 ) ⋅ 2 = lim x → 0 2 − ( x + 2 ) 2 x ( x + 2 ) = lim x → 0 − x 2 x ( x + 2 ) = lim x → 0 − 1 2 ( x + 2 ) = − 1 2 ( 0 + 2 ) = − 1 4

  $$\begin{align*}\lim_{x \to 0}\frac{(2+x)^{-1}-2^{-1}}{x}&=\lim_{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x} \cdot \frac{(x+2) \cdot 2}{(x+2) \cdot 2} \\ &=\lim_{x \to 0}\frac{2-(x+2)}{2x(x+2)} \\ &=\lim_{x \to 0}\frac{-x}{2x(x+2)} \\ &=\lim_{x \to 0}\frac{-1}{2(x+2)} \\ &=-\frac{1}{2(0+2)} \\ &=-\frac{1}{4} \end{align*}$$
  ::立方公尺xxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxx2+2-1-1xx=limxxxxxxxx01x+2-12xxxxxxxxxxxxx2xxxxxxxx=lxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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  Example 6
  ::例6

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  Evaluate the following limit:

  $$\begin{align*}\lim_{x \to -3}\frac{\sqrt{x^2-5}-2}{x+3}.\end{align*}$$
  ::评估以下限值:limx%3x2-5-2x+3。

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  Solution :
  ::解决方案 :

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  lim x → − 3 √ x 2 − 5 − 2 x + 3 = lim x → − 3 ( √ x 2 − 5 − 2 ) ( x + 3 ) ⋅ ( √ x 2 − 5 + 2 ) ( √ x 2 − 5 + 2 ) = lim x → − 3 ( x 2 − 5 − 4 ) ( x + 3 ) ( √ x 2 − 5 + 2 ) = lim x → − 3 ( x 2 − 9 ) ( x + 3 ) ( √ x 2 − 5 + 2 ) = lim x → − 3 ( x − 3 ) ( x + 3 ) ( x + 3 ) ( √ x 2 − 5 + 2 ) = lim x → − 3 x − 3 √ x 2 − 5 + 2 = − 3 − 3 √ 9 − 5 + 2 = − 6 2 + 2 = − 6 4 = − 3 2

  $$\begin{align*}\lim_{x \to -3}\frac{\sqrt{x^2-5}-2}{x+3}&=\lim_{x \to -3}\frac{\left(\sqrt{x^2-5}-2\right)}{(x+3)} \cdot \frac{\left(\sqrt{x^2-5}+2\right)}{\left(\sqrt{x^2-5}+2\right)} \\ &=\lim_{x \to -3}\frac{(x^2-5-4)}{(x+3)\left(\sqrt{x^2-5}+2\right)} \\ &=\lim_{x \to -3}\frac{(x^2-9)}{(x+3)\left(\sqrt{x^2-5}+2\right)} \\ &=\lim_{x \to -3}\frac{(x-3)(x+3)}{(x+3)\left(\sqrt{x^2-5}+2\right)} \\ &=\lim_{x \to -3}\frac{x-3}{\sqrt{x^2-5}+2} \\ &=\frac{-3-3}{\sqrt{9-5}+2} \\ &=\frac{-6}{2+2} \\ &=\frac{-6}{4} =\frac{-3}{2}\end{align*}$$
  ::3x2 - 5x2 - 5x3= limx3x3x3x3x3x3x3x3xx3xx3xxx2- 5(x2-5+2)(x2-5+2)= limx3x3x5x5+3)(x2-5+3)(x2-5+2)=limx3x3x2x9(x3+3)(x2-5+2)=limx3x3x3(x3+3)(x2+5+2)=limx3x3x3x3x2-x2(5+2)x3-3x-9(5+2+5)=lx3x3-3-9(x2+2)= 62+2_642+642***

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  Example 7
  ::例7

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  Evaluate the following limit:  

  $$\begin{align*}\lim_{x \to 0}\left( \frac{3}{x\sqrt{9-x}}-\frac{1}{x} \right).\end{align*}$$
  ::评估以下限值: limx%0( 3x%9- x- 1x) 。

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  Solution :
  ::解决方案 :

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  lim x → 0 ( 3 x √ 9 − x − 1 x ) = lim x → 0 ( 3 x √ 9 − x − √ 9 − x x √ 9 − x ) = lim x → 0 ( 3 − √ 9 − x x √ 9 − x ) = lim x → 0 ( ( 3 − √ 9 − x ) x √ 9 − x ⋅ ( 3 + √ 9 − x ) ( 3 + √ 9 − x ) ) = lim x → 0 ( 9 − ( 9 − x ) x √ 9 − x ( 3 + √ 9 − x ) ) = lim x → 0 x x √ 9 − x ( 3 + √ 9 − x ) = lim x → 0 1 √ 9 − x ( 3 + √ 9 − x ) = 1 √ 9 ( 3 + √ 9 ) = 1 3 ( 3 + 3 ) = 1 3 ( 6 ) = 1 18

  $$\begin{align*}\lim_{x \to 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x} \right) &=\lim_{x \to 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{\sqrt{9-x}}{x \sqrt{9-x}} \right) \\ &=\lim_{x \to 0}\left(\frac{3- \sqrt{9-x}}{x \sqrt{9-x}} \right) \\ &=\lim_{x \to 0}\left(\frac{\left(3- \sqrt{9-x}\right)}{x \sqrt{9-x}} \cdot \frac{\left(3+ \sqrt{9-x}\right)}{\left(3+ \sqrt{9-x}\right)} \right) \\ &=\lim_{x \to 0}\left(\frac{9-(9-x)}{x \sqrt{9-x} \left(3+ \sqrt{9-x}\right)} \right) \\ &=\lim_{x \to 0}\frac{x}{x \sqrt{9-x} \left(3+ \sqrt{9-x}\right)} \\ &=\lim_{x \to 0}\frac{1}{\sqrt{9-x} \left(3+ \sqrt{9-x}\right)} \\ &=\frac{1}{\sqrt{9}\left(3+ \sqrt{9}\right)} \\ &=\frac{1}{3(3+3)} \\ &=\frac{1}{3(6)} \\ &=\frac{1}{18}\end{align*}$$
  :3x9x9-x-1x) = limx0-0(3x9-9-x9-x) = limx0(3x9-xx9-9-x) = limx0(3x9-x) 9-x) = 3x9-x(9-x)(9-x)(3x9-x)(3x9-x)(3x) = 5xx9-9xx(3-x) = 5x0x0x(9-x) = 3x0x-9-x(9-x) = 1/9x(3-9) = 13(3+3) = 13(6) =118

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  Summary
  ::摘要

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  • Rationalization is a technique used to evaluate limits 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。
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  • With rationalization, you make the numerator and the denominator of an expression rational by using the properties of conjugate pairs.
    ::有了合理化,你就可以使用同性配对的特性, 使一个表达式的分子和分母合理化。

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  Review
  ::回顾

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

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

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  2. $\begin{align*}\lim_{x \to 4}\frac{\sqrt{x}-2}{x-4}\end{align*}$
  ::2. 脂质4x-2x-4

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

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

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  5. $\begin{align*}\lim_{x \to 4}\frac{\sqrt{3x+4}-x}{4-x}\end{align*}$
  ::5. 立度=4x3x+4-x4-x4-xx

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  6. $\begin{align*}\lim_{x \to 0}\frac{2-\sqrt{x+4}}{x}\end{align*}$
  ::6. limx02x+4x

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

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  8. $\begin{align*}\lim_{x \to 16}\frac{16-x}{4-\sqrt{x}}\end{align*}$
  ::8. 立方公尺1616-x4x

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  9. $\begin{align*}\lim_{x \to 0}\frac{x^2}{\sqrt{x^2+12}-\sqrt{12}}\end{align*}$
  ::9. 升=0x2x2x2+12+12=12

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  10. $\begin{align*}\lim_{x \to 2}\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2}\end{align*}$
  ::10. 立方公尺xxx+5xx+7x-2

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  11. $\begin{align*}\lim_{x \to 1}\frac{1-\sqrt{x}}{1-x}\end{align*}$
  ::11. 立方公尺

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

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

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  14. $\begin{align*} \lim_{x \to 9}\frac{9x^2-90x+81}{9-3\sqrt{x}}\end{align*}$
  ::14. 99x2-90x+819-3xx

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  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?

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  Review (Answers )
  ::回顾(答复)

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  Please see the Appendix.
  ::请参看附录。