14.4 替代确定限额

Section outline

• Select section General

  Collapse Expand

  General

  Collapse all Expand all

  • Select activity 新闻通告

    

    新闻通告 Forum
• Select section 替代查找限额

  Collapse Expand

  替代查找限额

  播放段落

  Finding limits for the vast majority of points for a given function is as simple as substituting the number that  $\begin{array}{r}x\end{array}$ 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. How can you decide if substitution is an appropriate analytical tool for finding a limit?
  ::特定函数绝大多数点的查找限制与替换 x 函数中选择的数值一样简单。 由于此选项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。 您如何决定替换是否是找到限制的适当分析工具?

  播放段落

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

  播放段落

  Finding a limit analytically means finding the limit using algebraic means. In order to evaluate many limits, you can substitute the value that  $\begin{array}{r}x\end{array}$ approaches into the function and evaluate the result. This works perfectly when there are no holes or asymptotes at that particular  $\begin{array}{r}x\end{array}$ value. You can be confident this method works as long as you don’t end up dividing by zero when you substitute.
  ::在分析上找到限制意味着使用代数手段找到限制。 为了评估许多限制, 您可以将 x 的值替换到函数中, 并评价结果 。 如果在特定的 x 值上没有空洞或无症状, 这个方法会非常有效 。 只要您在替换时不会以零除以零, 您可以相信这个方法有效 。

  播放段落

  If the function  $\begin{array}{r}f(x)\end{array}$ has no holes or asymptote at  $\begin{array}{r}x=a\end{array}$ then: $\begin{array}{r}\underset{x\to a}{lim}f(x)=f(a)\end{array}$
  ::如果函数 f( x) 在 x=a 时没有洞或无孔或无音效,则函数 f( x) =f( a) : limx_af( x) =f( a)

  播放段落

  Occasionally there will be a hole at $\begin{array}{r}x=a\end{array}$ . The limit in this case is the height of the function if the hole did not exist. In other words, if the function is a rational expression with factors that can be canceled, cancel the term algebraically and then substitute into the resulting expression. If no factors can be canceled, it could be that the limit does not exist at that point due to asymptotes.
  ::偶尔在 x=a 处会有一个洞。 在此情况下的限制是, 如果该洞不存在, 函数的高度。 换句话说, 如果该函数是一个理性表达式, 且含有可以取消的因素, 则取消代数, 然后替换为生成的表达式。 如果无法取消任何因素, 则该限制可能因微调而不存在 。

  播放段落

  Look at the following two limits:
  ::看看以下两个限制:

  播放段落

  $\begin{array}{r}\underset{x\to 2}{lim}\frac{x^2-4}{x-2},\underset{x\to 3}{lim}\frac{x^2-4}{x-2}\end{array}$
  ::2x2 -4x-2, limx3x2 -4x-2, limx3x2 - 4x-2

  播放段落

  The limit on the left cannot be evaluated by direct substitution because if 2 is substituted in, then you end up dividing by zero. The limit on the right can be evaluated using direct substitution because the hole exists at  $\begin{array}{r}x=2\end{array}$ not $\begin{array}{r}x=3\end{array}$ . Thus, the limit is:
  ::左侧的限值无法通过直接替代来评估, 因为如果将 2 替换为 2, 那么最终会除以 零 。 右侧的限值可以通过直接替代来评估, 因为洞位位于 x= 2 而不是 x= 3 。 因此, 限值是 :

  播放段落

  $\begin{array}{r}\underset{x\to 3}{lim}\frac{x^2-4}{x-2}=\frac{3^2-4}{3-2}=\frac{9-4}{1}=5\end{array}$
  ::3x2-4x-2=32-43-2=9-41=5

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked how to determine if you should use substitution to solve a limit. In order to decide whether substitution is an appropriate first step you can always just try it. You’ll know it won’t work if you end up trying to evaluate an expression with a denominator equal to zero.  If this happens, go back and try to factor and cancel, and then try substituting again.
  ::早些时候,有人问您如何确定您是否应该使用替代来解决限制问题。 为了决定替换是否是一个适当的第一步,您总可以尝试一下。 如果最终试图用等于零的分母来评价表达式,你就会知道它不会起作用。 如果发生这种情况,你回去考虑一下,然后取消,然后再尝试替换。

  播放段落

  Example 2
  ::例2

  播放段落

  Evaluate the following limit by canceling first and then using substitution.
  ::通过首先取消,然后使用替代方法,对以下限制进行评估。

  播放段落

  $\begin{array}{r}\underset{x\to 2}{lim}\frac{x^2-4}{x-2}\end{array}$
  ::limx% 2x2 - 4x- 2

  播放段落

  $$\begin{array}{rl}\underset{x\to 2}{lim}\frac{x^2-4}{x-2}& =\underset{x\to 2}{lim}\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)}\\ & =\underset{x\to 2}{lim}(x+2)\\ & =2+2\\ & =4\end{array}$$

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

  播放段落

  Example 3
  ::例3

  播放段落

  Evaluate the following limit analytically:  $\begin{array}{r}\underset{x\to 4}{lim}\frac{x^2-x-12}{x-4}\end{array}$
  ::分析评估以下限度: limx4x2-x-12x-4。

  播放段落

  $$\begin{array}{rl}\underset{x\to 4}{lim}\frac{x^2-x-12}{x-4}& =\underset{x\to 4}{lim}\frac{\left(x-4\right)\left(x+3\right)}{\left(x-4\right)}\\ & =\underset{x\to 4}{lim}(x+3)\\ & =4+3\\ & =7\end{array}$$

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

  播放段落

  Example 4
  ::例4

  播放段落

  Evaluate the following limit analytically.
  ::以分析方式评估以下限制。

  播放段落

  $\begin{array}{r}\underset{x\to 3}{lim}\frac{x^2-9}{x-3}\end{array}$
  ::limx=3x2-9x-3

  播放段落

  $\begin{array}{r}\underset{x\to 3}{lim}\frac{x^2-9}{x-3}=\underset{x\to 3}{lim}\frac{\left(x-3\right)\left(x+3\right)}{\left(t-3\right)}=\underset{x\to 3}{lim}(x+3)=6\end{array}$
  ::立方厘米x3x2-9x-3=limx3-3(x-3)(x+3)(t-3)(t-3)=limx3x3(x+3)=6

  播放段落

  Example 5
  ::例5

  播放段落

  Evaluate the following limit analytically.
  ::以分析方式评估以下限制。

  播放段落

  $\begin{array}{r}\underset{y\to 4}{lim}\frac{3|y-1|}{y+4}\end{array}$
  ::

  播放段落

  $\begin{array}{r}\underset{y\to 4}{lim}\frac{3|y-1|}{y+4}=\frac{3|4-1|}{4+4}=\frac{3\cdot 3}{8}=\frac{9}{8}\end{array}$
  ::=34 -14+4=34 -14+4=338=98

  Summary

  播放段落 Finding limits analytically involves using algebraic means, such as substitution. ::在分析上,发现极限涉及使用代数手段,如替代。 播放段落 Substitution works when there are no holes or asymptotes at the value being approached. ::当所接触的数值没有孔或无症状时,替代就起作用。 播放段落 Be cautious of dividing by zero when using substitution to find limits. ::在使用替代物以寻找限制时,要谨慎小心,不要以零除以零。 播放段落 If there is a hole at the value being approached, cancel the term algebraically and then substitute into the resulting expression. ::如果所接触的值存在空洞, 取消代数术语, 然后替换为结果表达式 。 播放段落 If no factors can be canceled, the limit may not exist at that point due to asymptotes. ::如果无法取消任何因素,那么由于微粒,此时的限制可能不存在。

  播放段落

  Review
  ::回顾

  播放段落

  Evaluate the following limits analytically.
  ::分析评估以下限度。

  播放段落

  1. $\begin{array}{r}\underset{x\to 5}{lim}\frac{x^2-25}{x-5}\end{array}$
  ::1. limx=5x2-2-25x-5

  播放段落

  2. $\begin{array}{r}\underset{x\to -1}{lim}\frac{x^2-3x-4}{x+1}\end{array}$
  ::2. limx%1x2-3x-4x+1

  播放段落

  3. $\begin{array}{r}\underset{x\to 5}{lim}\sqrt{5x}-12\end{array}$
  ::3. limx55x-12

  播放段落

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

  播放段落

  5. $\begin{array}{r}\underset{x\to 1}{lim}\frac{3x|x-4|}{x+1}\end{array}$
  ::5. 立方公尺13xx-4x+1

  播放段落

  6. $\begin{array}{r}\underset{x\to 2}{lim}\frac{x^2+5x-14}{x-2}\end{array}$
  ::6. limx%2x2+5x-14x-2

  播放段落

  7. $\begin{array}{r}\underset{x\to 1}{lim}\frac{x^2-8x+7}{x-1}\end{array}$
  ::7. limx=1x2-8x+7x-1

  播放段落

  8. $\begin{array}{r}\underset{x\to 0}{lim}\frac{5x-1}{2x^2+3}\end{array}$
  ::8. limx05x-12x2+3

  播放段落

  9. $\begin{array}{r}\underset{x\to 1}{lim}4x^2-2x+5\end{array}$
  ::9. limx}14x2-2x+5

  播放段落

  10. $\begin{array}{r}\underset{x\to 0}{lim}\frac{x^2+5x}{x}\end{array}$
  ::10. limx=0x2+5xxx

  播放段落

  11. $\begin{array}{r}\underset{x\to -3}{lim}\frac{x^2-9}{x+3}\end{array}$
  ::11. limx=%3x2-9x+3

  播放段落

  12. $\begin{array}{r}\underset{x\to 0}{lim}\frac{5x+1}{x}\end{array}$
  ::12. limx05x+1x

  播放段落

  13. $\begin{array}{r}\underset{x\to 1}{lim}\frac{5x+1}{x}\end{array}$
  ::13. limx=15x+1x

  播放段落

  14. $\begin{array}{r}\underset{x\to 5}{lim}\frac{x^2-25}{x^3-125}\end{array}$
  ::14. 立方×5x2-2-25x3-125

  播放段落

  15. $\begin{array}{r}\underset{x\to -1}{lim}\frac{x-2}{x+1}\end{array}$
  ::15. limx%1x-2x+1

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。