8.4 多边作用限度

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  多元函数极限

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  You may recall that many of the rules you have learned to use when solving basic algebra equations often apply to the manipulation of complex equations also. You will find in this lesson that a similar concept applies to operations on limits.
  ::你也许记得,在解决基本代数方程式时学会使用的许多规则也常常适用于复杂方程式的操纵。 从这一教训中,你可以看到类似的概念适用于有限度的操作。

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  Polynomial Function Limits
  ::多元函数极限

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  In this lesson, you will evaluate the limits of functions from an algebraic perspective. First you will require the proper tools for the job, the theorems necessary to calculate limits. These theorems are outlined in the two boxes below.
  ::在此课程中, 您将从代数角度评估函数的限度。 首先, 您需要合适的工具, 计算限制所需的理论。 这些理论将在下面的两个框中概述 。

  播放段落 Box #1: Important Theorems of Limits ::框 # 1: 重要界限的理论 播放段落 Let a be a real number and suppose that $\begin{align*}\lim_{x\rightarrow a}f(x) = L_1 \end{align*}$ and $\begin{align*}\lim_{x\rightarrow a}g(x) = L_2 \end{align*}$ . ::假設 limxaf(x) = L1 和 limxag(x) = L2 。 播放段落 Then: ::然后: 播放段落 1. $\begin{align*} \lim_{x\rightarrow a} [f(x) + g(x)] = \lim_ {x\rightarrow a}f(x) + \lim_{x\rightarrow a}g(x) = L_1 + L_2\end{align*}$ , meaning the limit of the sum is the sum of the limits. ::1. limxa[f(x)+g(x)]=limxaf(x)+limxag(x)=L1+L2,这意味着总和的上限是限额的和。 播放段落 2. $\begin{align*} \lim_{x\rightarrow a} [f(x) - g(x)] = \lim_ {x\rightarrow a}f(x) - \lim_{x\rightarrow a}g(x) = L_1 - L_2\end{align*}$ , meaning the limit of the difference is the difference of the limits. ::2. limxa[f(x)-g(x)]=limxaf(x)-limxag(x)=L1-L2,即差值的限度是限额的差数。 播放段落 3. $\begin{align*} \lim_{x\rightarrow a} [f(x)g(x)] = ( \lim_ {x\rightarrow a}f(x)) ( \lim_{x\rightarrow a}g(x)) = L_1L_2\end{align*}$ , meaning the limit of the product is the product of the limits. ::3. limxa[f(x)g(x)]=(limxaf(x))(limxag(x))=L1L2,这意味着产品的限值是限值的产物。 播放段落 4. $\begin{align*} \lim_{x\rightarrow a} \frac{f(x)}{g(x)} =\frac{\lim_{x\rightarrow a} f(x)} {\lim_{x\rightarrow a} g(x)} = \frac {L_1} {L_2} L_2 \neq 0\end{align*}$ , meaning the limit of a quotient is the quotient of the limits (provided that the denominator does not equal zero.) ::4. limxaf(x)g(x)=limxaf(x)limxag(x)=L1L2L20,意指商数的限值是限值的商数(前提是分母不等于零)。 播放段落 5. $\begin{align*} \lim_{x\rightarrow a} \sqrt[n]{f(x)}= \sqrt[n] { \lim_{x\rightarrow a} f(x)} = \sqrt[n] L_1 L_1 > 0\end{align*}$ if n is even, meaning the limit of the n th root is the n th root of the limit. ::5. limxanf(x)=nlimxaf(x)=nL1L1>0,如果 n是偶数,表示 nth根的限值是限制的 nth根 。

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  Other useful results follow from the above theorems:
  ::从上述理论中得出的其他有益成果如下:

  播放段落 Box #2 ::方框2 播放段落 1. If a and k are real numbers, then $\begin{align*}\lim_{x\rightarrow a}k = k\end{align*}$ . That is, if f ( x ) = k , a constant function, then the values of f ( x ) do not change as x is varied, thus the limit of f ( x ) is k . ::1. 如果a和k是真实数字,则Limxak=k。也就是说,如果f(x)=k,则f(x)=k是一个常数函数,那么f(x)的值不会随着x的变化而变化,因此f(x)的限值是k。 播放段落 2. If a is a real number then $\begin{align*}\lim_{x\rightarrow a} x= a \end{align*}$ . That is, since f ( x ) = x is an identity function (its input equals its output), then as x → a , f ( x ) = x → a . ::2. 如果a是一个实际数字,那么limxax=a。也就是说,f(x)=x是一个身份函数(其输入等于其输出),然后作为 x a, f(x)=x a。 播放段落 3. $\begin{align*}\lim_{x \rightarrow a} (k \cdot f(x)) = ( \lim_{x \rightarrow a} k) \cdot (\lim_{x \rightarrow a} f(x)) = k \cdot ( \lim_{x \rightarrow a} f(x))\end{align*}$ ::3. 立方a(kf(x))=(limxak) (limxaf(x)) =k(limxaf(x) ) 播放段落 4. $\begin{align*}\lim_{x \rightarrow a} x^n = ( \lim_{x \rightarrow a} x)^n = a^n\end{align*}$ ::4. 立方xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxn=an

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  Finally, we have one more theorem:
  ::最后,我们还有一个理论:

  播放段落 Theorem: The limit of a polynomial ::定理: 多式理论的极限 For any polynomial f ( x ) = c n x n + . . . + c 1 x + c 0 and any real number a , $\begin{align*}\lim_{x \rightarrow a}f(x) = c_n(a)^n + . . . + c_1(a) + c_0\end{align*}$ $\begin{align*}\lim_{x \rightarrow a}f(x) = f(a)\end{align*}$ In other words, the limit of the polynomial is simply equal to f ( a ).

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  Examples
  ::实例

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  Example 1
  ::例1

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  Use the theorems above to find $\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4)\end{align*}$ and justify each step.
  ::使用上面的定理查找 limx1(x2- 3x+4) 并解释每个步骤的理由 。

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  You can verify your practice application of the theorems described above by simply substituting x = 1 directly into the polynomial.
  ::您可以通过将 x = 1 直接替换为多元体来验证上述理论的实际应用情况。

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  Using Equation (1) of Box #1 (the limit of the sum is the sum of the limits) we get
  ::使用方框1的等量(1)(总和的限度是限额的和),我们得到的

  	$\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = \lim_{x \rightarrow 1}x^2 + \lim_{x \rightarrow 1}(-3x) + \lim_{x \rightarrow 1} 4\end{align*}$

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  From Equation (4) of Box #2, the first term becomes
  ::从方框2的等式(4)中,第一学期改为

  	$\begin{align*}\lim_{x \rightarrow 1}x^2 = (1)^2 = 1\end{align*}$

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  From Equations (2) and (3) of Box #2, the second term becomes
  ::从方框2中的等数(2)和(3)中,第二任期改为

  	$\begin{align*}\lim_{x \rightarrow 1}(-3x) = -3 \lim_{x \rightarrow 1}x = (-3)(1) = -3\end{align*}$

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  Finally, from Equation (1) of Box #2, the third term becomes
  ::最后,从方框2的等式(1)中,第三学期成为

  	$\begin{align*}\lim_{x \rightarrow 1}4 = 4\end{align*}$

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  Thus the limit of the above polynomial is
  ::因此,上述多元性的限制是:

  	$\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = 1 + (-3) + 4 = 2\end{align*}$

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  Example 2
  ::例2

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  Find $\begin{align*}\lim_{x\rightarrow 3} (4x^3 - 4x - 5)\end{align*}$ .
  ::查找 limx%3( 4x3- 4x- 5) 。

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  According to the theorem above, $\begin{align*}\lim_{x\rightarrow 3}(4x^3 - 4x - 5) = 4(3)^3 - 4(3) - 5 =91\end{align*}$
  ::根据上面的理论, limx}3( 4x3- 4x- 5) = 4(3)-3-4(3)-5=91

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  Example 3
  ::例3

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  Find $\begin{align*}\lim_{x\rightarrow 5} \frac{x^2 - 4} {3x^2 - 2} \end{align*}$ .
  ::查找 limx%5x2 - 43x2 - 2 。

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  Using Equation (4) of Box #1 (the limit of the quotient is the quotient of the limit),
  ::使用方框1的等式(4)(商数的限度是限度的商数),

  	$\begin{align*}\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2} = \frac {\lim_{x \rightarrow 5} (x^2 - 4)} {\lim_{x \rightarrow 5}(3x^2 - 2)}\end{align*}$

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  Making use of the limit of the polynomials theorem, we obtain
  ::利用多元论定理的极限,我们获得

  	$\begin{align*}\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2}\end{align*}$	$\begin{align*}= \frac{(5)^2 - 4} {3(5)^2 - 2}\end{align*}$
  		$\begin{align*}\mathit = \frac{21} {73}\end{align*}$

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  Example 4
  ::例4

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  Find the limit: $\begin{align*}\lim_{x \to -3} \left( \frac{x^2 + 3x}{x + 3}\right)\end{align*}$ .
  ::查找限制值: limx%3( x2+3xx+3) 。

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  $\begin{align*}\left( \frac{x(x + 3)}{x + 3}\right)\end{align*}$ ..... First factor the numerator
  :x( x+3) x+3)...。第一个乘数

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  Since both numerator and denominator contain $\begin{align*}(x + 3)\end{align*}$ the function output is the same as the input for any number except where $\begin{align*}(x + 3)\end{align*}$ was undefined, -3.
  ::由于分子和分母都包含(x+3),函数输出与任何数字的输入相同,除非(x+3)未定义, -3。

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  Since x = y for any number other than -3, as we get closer and closer to -3 on the input, we get closer and closer to -3 on the output (we can get as close as we like, we just can't hit it!).
  ::由于 X = y 表示除 - 3之外的任何数字, 当我们接近输入的 - 3时, 我们接近 - 3 表示输出的 - 3 时( 我们可以尽可能接近, 我们不能击中它 ! ) !

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  $\begin{align*}\therefore \lim_{x \to -3} \left( \frac{x^2 + 3x}{x + 3}\right) = -3\end{align*}$
  ::3(x2+3xx+3)3

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  Example 5
  ::例5

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  Given:
  ::参照:

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  $\begin{align*}\lim_{x \to c} f(x) = 7\end{align*}$
  ::limxcf(x)=7

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  $\begin{align*}\lim_{x \to c} g(x) = 3\end{align*}$
  ::立方厘米( x)=3

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  Find: $\begin{align*}\lim_{x \to c} f(x) g(x)\end{align*}$
  ::查找: limxcf(x)g(x)

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  This example uses rule #3 from above: the limit of the product is the product of the limits. Which means that we need to find the limit of each function, then multiply the limits.
  ::此示例使用上面的规则3: 产品的限值是限值的产物。 这意味着我们需要找到每个函数的限值, 然后乘以限值 。

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  $\begin{align*}\lim_{x \to c} f(x) = 7\end{align*}$
  ::limxcf(x)=7

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  $\begin{align*}\lim_{x \to c} g(x) = 3\end{align*}$
  ::立方厘米( x)=3

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  Since we are given the limits, we simply multiply $\begin{align*}7 \cdot 3 = 21\end{align*}$
  ::既然我们得到了极限,我们只需乘以73=21

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  $\begin{align*}\therefore \lim_{x \to c} f(x) g(x) = 21\end{align*}$
  ::x=21=xxxxxx=21=xxxxxxxxxxxxxxxxxx=21=xxxxxxxxxxxxxxxxxxxxxxxxxxx

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  Example 6
  ::例6

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  Given:
  ::参照:

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  $\begin{align*}f(x) = 4x^2 + 3x + 4\end{align*}$
  ::f(x)=4x2+3x+4

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  $\begin{align*}g(x) = 2x^2 -5x -2\end{align*}$
  ::g(x) = 2x2 - 5x-2

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  Find: $\begin{align*}\lim_{x \to -2} f(g(x))\end{align*}$
  ::查找: limx%%2f( g( x) )

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  Since these are both continuous functions, we can use $\begin{align*}\lim f(g(x)) = f \left(\lim g(x)\right)\end{align*}$
  ::由于这两个功能都是连续的功能,我们可以使用limf(g(x))=f(limg(x))

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  $\begin{align*}\lim_{x \to -2} 2x^2 -5x -2 = 16\end{align*}$
  ::limx22x2 - 5x-2=16

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  $\begin{align*}f(16) = 4(16)^2 + 3(16) + 4 = 1076\end{align*}$
  ::f(16)=4(16)2+3(16)+4=1076

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  $\begin{align*}\therefore \lim_{x \to -2} f(g(x)) = 1076\end{align*}$
  ::2f( g( x)) =1076

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  Review
  ::回顾

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  Find the limit:
  ::查找限制 :

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  1. $\begin{align*}\lim_{x \to 2} -5x - 2= \end{align*}$
     ::limx%2 - 5x-2=
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  2. $\begin{align*}\lim_{x \to \frac{\pi}{6}} csc (x) = \end{align*}$
     ::limx6csc(x)=
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  3. $\begin{align*}\lim_{x \to 256} \sqrt[4]{x}= \end{align*}$
     ::立方厘米=2564x=
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  4. $\begin{align*}\lim_{x \to0 } \frac{x^2 - 2x}{x}= \end{align*}$
     ::limx=0x2-2xxx=
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  5. $\begin{align*}\lim_{x \to 1} 4x = \end{align*}$
     ::limx=14x=

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  Find the limit:
  ::查找限制 :

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  6. $\begin{align*}\lim_{x \to -2}\sqrt{-3x + 3} = \end{align*}$
     ::======================================================================================================================================================================================================
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  7. Given: $\begin{align*}\lim_{x \to c} f(x) = -1 \end{align*}$ and $\begin{align*}\lim_{x \to c } g(x) = -1\end{align*}$ , find: $\begin{align*}\lim_{x \to c} f(x) - g(x) = \end{align*}$
     ::参考文献: limx*cf(x)%1 和 limx*cg(x)%1, 查找: limx*cf(x)-g(x)=
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  8. Given: $\begin{align*}f(x) = -2x^2 + x - 3\end{align*}$ and $\begin{align*}g(x) = -x + 2\end{align*}$ , find: $\begin{align*}\lim_{x \to -5} f(g(x))= \end{align*}$
     ::给定 : f( x) 2x2+x-3 和 g( x) x+2, 找到: limx=5f( g( x)) =
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  9. Given: $\begin{align*}\lim{x \to c} f(x) = -2 \end{align*}$ and $\begin{align*}\lim_{x \to c} g(x) = -5 \end{align*}$ , find: $\begin{align*}\lim_{x \to c} f(x) + g(x)= \end{align*}$
     ::参考文献: limx*cf(x)%2 和 limx*cg(x)%5, 参见: limx*cf(x)+g(x)=
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  10. Given: $\begin{align*}f(x) = 2x^2 - 2x + 3\end{align*}$ and $\begin{align*}g(x) = -3x^2 - 4x - 5\end{align*}$ , find: $\begin{align*}\lim_{x \to 0} = g(f(x))=\end{align*}$
      ::根据: f( x) = 2x2 - 2x+3 和 g( x) = 3x2 - 4x-5, 找到: limx= 0= g( f( x)) =
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  11. $\begin{align*}\lim_{x \to-1} \sqrt{-5x^3 - 5x^2 - 5x - 3} = \end{align*}$
      ::5x3 - 5x2 - 5x3=
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  12. Given: $\begin{align*}\lim_{x \to c} f(x) = 7 \end{align*}$ and $\begin{align*}\lim_{x \to c} g(x) = -3 \end{align*}$ , find: $\begin{align*}\lim_{x \to c} f(x) - g(x) = \end{align*}$
      ::参考文献: limx=7 和 limx=cg=3, 参见: limx=x-g(x)=
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  13. Given: $\begin{align*}f(x) = x^2 - 2x + 2\end{align*}$ and $\begin{align*}g(x) = 4x^2 - 2x - 2\end{align*}$ , find: $\begin{align*}\lim_{x \to-1 }f(g(x)) = \end{align*}$
      ::根据: f( x) =x2-2x+2 和 g( x) = 4x2-2x-2, 查找: limx=1f( g( x)) =
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  14. $\begin{align*}\lim_{x \to 0} \sqrt{-4x^2 + 4x - 3}= \end{align*}$
      ::立方公尺xx2+4x3=
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  15. Given: $\begin{align*}\lim_{x \to c} f(x) = 5 \end{align*}$ and $\begin{align*}\lim_{x \to c} g(x) = 8\end{align*}$ , find: $\begin{align*}\lim_{x \to c} f(x) + g(x)= \end{align*}$
      ::依据: limx=5 和 limx=8, 参见: limx=x+g(x) =
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  16. Given: $\begin{align*}f(x) = -x^2 -3x -4 \end{align*}$ and $\begin{align*}g(x) = 2x^2 - 5x + 2\end{align*}$ , find: $\begin{align*}\lim_{x \to 4} f(g(x)) = \end{align*}$
      ::根据: f( x) x2- 3x-4 和 g( x) = 2x2 - 5x+2, 查找: limx4f( g( x)) =
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  17. $\begin{align*}\lim_{x \to-6 } \sqrt{-5x - 4} = \end{align*}$
      ::=================================================================================================================================================== ================================================================================================================

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  Review (Answers)
  ::回顾(答复)

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  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 。