2.6 涉及激进职能的限制

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 涉及激进职能的限制

  Collapse Expand

  涉及激进职能的限制

  播放段落

  There are many problems that will involve taking the $\begin{array}{r}n\end{array}$ th root of a variable expression, so it is natural that there may sometimes be a need to find the limit of a function involving radical expressions, using square or cube roots, or other roots. Do you think that finding the limit of a function involving radicals would be any different than finding the limit of polynomial or rational functions? Can you think of any ways that radicals might present different problems than polynomials?
  ::有许多问题需要从变量表达式的 nth 根中找到一个变量表达式的nth 根,因此,有时可能需要找到一个函数的极限,这个函数涉及激进表达式,使用正方根或立方根或其他根。 你认为找到一个函数的极限与找到多元函数或理性函数的界限有什么不同吗? 你能想象到激进分子可能会产生与多元函数不同的问题吗?

  播放段落

  Limits with Radical Functions
  ::带有激进功能的限制

  播放段落

  When evaluating a limit involving a radical function , use direct substitution to see if a limit can be evaluated whenever possible. If not, other methods to evaluate the limit need to be explored.
  ::在评估涉及激进功能的限制时,应使用直接替代方法,以确定是否可以尽可能对限制进行评价,否则,需要探索其他方法来评估限制。

  播放段落

  Take the following function   $\begin{array}{r}f(x)=\sqrt{x}-3\end{array}$ . Find  $\begin{array}{r}\underset{x\to 9}{lim}\sqrt{x}-3\end{array}$ . 
  ::采取以下函数 f( x) =x- 3. 查找 limx=% 9x- 3 。

  播放段落

  $$\begin{array}{rl}\underset{x\to 9}{lim}\sqrt{x}-3& =\underset{x\to 9}{lim}\sqrt{x}-\underset{x\to 9}{lim}3\\ \underset{x\to 9}{lim}\sqrt{x}-3& =\underset{x\to 9}{lim}\sqrt{x}-\underset{x\to 9}{lim}3\\ & =\sqrt{9}-3\\ & =0\end{array}$$

  
  ::limx_9x-3=limx_9x-limx_93limx_9x-3=limx_9x-limx_93=9-3=0

  播放段落

  Therefore, $\begin{array}{r}\underset{x\to 9}{lim}\sqrt{x}-3=0\end{array}$ , which could have been determined by directly evaluating  $\begin{array}{r}f(x)\end{array}$ at $\begin{array}{r}x=9\end{array}$ , i.e., by using direct substitution.
  ::因此, limx%9x-3=0, 可以通过直接评估x=9的f(x), 即使用直接替代来确定。

  播放段落

  Now, find  $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}\sqrt{x}-3\end{array}$ .
  ::现在,找到Limxxxx -3。

  播放段落

  Evaluating  $\begin{array}{r}f(x)\end{array}$ at ever increasing positive values of  $\begin{array}{r}x\end{array}$ shows that  $\begin{array}{r}f(x)\end{array}$ increase without bound. Therefore, $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}\sqrt{x}-3=\mathrm{\infty }\end{array}$ .
  ::在不断增长的x正值上评价 f(x) 显示 f(x) 增长无约束。 因此, limxx- 3\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ f(x)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

  播放段落

  In both of the above cases, direct substitution could be used to evaluate the limits and there is no need for alternative methods. 
  ::在上述两种情况下,直接替代都可用于评估限度,没有必要采用替代方法。

  播放段落

  Take a look at the function  $\begin{array}{r}g(x)=\frac{\sqrt{x^2+3}}{7x+5}\end{array}$ .   Find  $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x^2+3}}{7x+5}\end{array}$ .
  ::查看函数 g( x) =x2+37x+5 。 查找 limx\\ x2+37x+5 。

  播放段落

  First we notice that we should exclude $\begin{array}{r}x=-\frac{5}{7}\end{array}$  in any evaluation. Using direct substitution to find the limit results in the indeterminate form $\begin{array}{r}\frac{\mathrm{\infty }}{\mathrm{\infty }}\end{array}$ . To transform the radical expression to a better form, use the fact that the value of  $\begin{array}{r}x\end{array}$ is going to larger and larger positive values. This allows the following:
  ::首先,我们注意到,在任何评估中,我们都应该排除 x*%57。使用直接替代来找到不确定形式的限值结果 。要将激进表达方式转换为更好的形式,请使用 x 的值将变成更大和更大的正值这一事实。 允许以下内容 :

  播放段落

  $$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x^2+3}}{7x+5}& =\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x^2\left(1+\frac{3}{x^2}\right)}}{x\left(7+\frac{5}{x}\right)}\\ \underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x^2+3}}{7x+5}& =\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x^2\left(1+\frac{3}{x^2}\right)}}{x\left(7+\frac{5}{x}\right)}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{|x|\sqrt{\left(1+\frac{3}{x^2}\right)}}{x\left(7+\frac{5}{x}\right)}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}\sqrt{\left(1+\frac{3}{x^2}\right)}}{\underset{x\to \mathrm{\infty }}{lim}\left(7+\frac{5}{x}\right)}\\ & =\frac{1}{7}\end{array}$$

  
  ::limxxxx2+37x+5=limxxxxxxx2(1+3x2)x(7+5x)xxxxx+37x+5=limxxxxxxx2(1+3x2)x(7+5x)x(1+3x2)x(7+5x)x(7+5x)x(7+5x)x(1+3x2)x(7+5x)x(7+5x)x=17

  播放段落

  Therefore, $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x^2+3}}{7x+5}=\frac{1}{7}\end{array}$ .
  ::因此, limxx2+37x+5=17。

  播放段落

  Now, find  $\begin{array}{r}\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{x^2+3}}{7x+5}\end{array}$ .
  ::现在,找到limxx2+37x+5。

  播放段落

  The solution to evaluating the limit at negative infinity is similar to the above approach except that  $\begin{array}{r}x\end{array}$ is always negative.
  ::以负无限度评价限额的办法与上述办法相似,但x总是负的。

  播放段落

  $$\begin{array}{rl}\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{x^2+3}}{7x+5}& =\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{x^2\left(1+\frac{3}{x^2}\right)}}{x\left(7+\frac{5}{x}\right)}\\ & =\underset{x\to -\mathrm{\infty }}{lim}\frac{|x|\sqrt{\left(1+\frac{3}{x^2}\right)}}{x\left(7+\frac{5}{x}\right)}\\ & =\frac{\underset{x\to -\mathrm{\infty }}{lim}\sqrt{\left(1+\frac{3}{x^2}\right)}}{-\underset{x\to -\mathrm{\infty }}{lim}\left(7+\frac{5}{x}\right)}\dots \text{Note the denominator has a “-“ because}x<0\\ & =-\frac{1}{7}\end{array}$$

  
  ::limxxxx2+37x+5=limxxxxxxx2(1+3x2)x(7+5x)=limxxxx*(1+3x2)x(7+5x)x(7+5x)=limx *(1+3x2)-limx*(7+5x)...注意分母有一个“-”因为 x<0{17)

  播放段落

  Therefore, $\begin{array}{r}\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{x^2+3}}{7x+5}=-\frac{1}{7}\end{array}$ .
  ::因此, limxx2+37x+5+17。

  播放段落

  So far, you have been able to find the limit of rational functions using methods shown earlier. However, there are times when this is not possible. Take the function $\begin{array}{r}h(x)=\frac{\sqrt{x}-3}{x-9}\end{array}$ . Find  $\begin{array}{r}\underset{x\to 9}{lim}h(x)\end{array}$ .
  ::到目前为止,您已经能够使用前面显示的方法找到理性函数的极限。 但是, 在一定的时间里, 这是不可能的 。 使用函数 h( x) =x- 3x- 9. 查找 lix% 9h( x) 。

  播放段落

  Using direct substitution to find the limit results in the indeterminate form $\begin{array}{r}\frac{0}{0}\end{array}$ . In order to evaluate the limit, we need to transform the expression to remove the indeterminate form. This is accomplished by using the relationship for the difference of squares of real numbers: $\begin{array}{r}{x}^2-y^2=(x+y)(x-y)\end{array}$ .
  ::使用直接替代来在不确定的表格 00 中找到极限结果。 为了评估极限, 我们需要将表达式转换为删除不确定的形式。 要做到这一点, 需要使用真实数字的正方形( x2- y2= (x+y)( x- y)) 的差数来匹配 。

  播放段落

  We then rewrite and simplify the original function as follows:
  ::然后我们改写和简化原功能如下:

  播放段落

  $$\begin{array}{rl}\underset{x\to 9}{lim}\frac{\sqrt{x}-3}{x-9}& =\underset{x\to 9}{lim}\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\dots \text{Use the difference of squares factoring to remove the 0 in the denominator.}\\ & =\underset{x\to 9}{lim}\frac{1}{(\sqrt{x}+3)}\\ & =\frac{1}{\underset{x\to 9}{lim}\left(\sqrt{9}+3\right)}\\ & =\frac{1}{6}\end{array}$$

  
  ::limx%9x-3x-9=limx%9x-3(x-3)(x+3)...使用方位乘数差来去除分母中的 0。=limx%91(x+3)=1limx9(9+3)=16

  播放段落

  Hence $\begin{array}{r}\underset{x\to 9}{lim}\frac{\sqrt{x}-3}{x-9}=\frac{1}{6}\end{array}$ .
  ::因此,9x-3x-9=16。

  播放段落

  Now, find the end behavior of the same function, i.e. find  $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}h(x)\end{array}$ .
  ::现在,找到相同函数的结束行为, 即找到 limxh( x) 。

  播放段落

  As  $\begin{array}{r}x\end{array}$ increases to large positive values, the function takes on the indeterminate form $\begin{array}{r}\frac{\mathrm{\infty }}{\mathrm{\infty }}\end{array}$ . The transformation above can also be used to evaluate the limit (Approach 1), as well as the technique used in evaluating rational functions (Approach 2).
  ::由于 x 增加为大正值, 函数以不确定的形式 。 以上变换也可以用来评价极限( Approach 1) , 以及评价合理函数所使用的技术( Approach 2) 。

  播放段落

  $$\begin{array}{rlrl}& \mathbf{A}\mathbf{p}\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{a}\mathbf{c}\mathbf{h}1& & \mathbf{A}\mathbf{p}\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{a}\mathbf{c}\mathbf{h}2\\ \underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x}-3}{x-9}& =\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}& & =\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x}\left(1-\frac{3}{\sqrt{x}}\right)}{x\left(1-\frac{9}{x}\right)}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{1}{(\sqrt{x}+3)}& & =\underset{x\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{x}}\cdot \underset{x\to \mathrm{\infty }}{lim}\frac{\left(1-\frac{3}{\sqrt{x}}\right)}{\left(1-\frac{9}{x}\right)}\\ & =\frac{1}{\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt{x}+3\right)}& & =0\cdot 1\\ & =0& & =0\end{array}$$

  
  ::Appropach 2limx*x-3x-9=limx*(x-3)(x+3)=limx*x(1-3x)x(1-9x)=limx**1(x+3)=limx*1(x+3)=lex*1(x+3)=(1-3x)(1-9x)=1limx*(x+3)=0=0=0

  播放段落

  Hence $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x}-3}{x-9}=0\end{array}$ .
  ::因此, limxxxx- 3x- 9=0 。

  播放段落

  Finally, find  $\begin{array}{r}\underset{x\to -\mathrm{\infty }}{lim}h(x)\end{array}$ .  The solution to this problem is that $\begin{array}{r}\underset{x\to -\mathrm{\infty }}{lim}h(x)\end{array}$   does not exist because the domain of  $\begin{array}{r}h(x)\end{array}$ does not include $\begin{array}{r}x<0\end{array}$ .
  ::最后,找到 limxh(x) 。 解决这个问题的解决方案是 limxh(x) 不存在, 因为 h(x) 的域不包括 x<0 。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked if the methods for evaluating limits involving polynomials and rational funct ions can be used to find the limits of radical functions. Some of the methods do work for radical functions. The use of direct substitution is a common method. Transforming indeterminate or undefined forms by finding and canceling common factors in the numerator and denominator, or factoring and simplifying the highest degree powers of variables represent common approaches.
  ::早些时候,有人问您,对于涉及多元性和理性功能的界限,是否可以使用评估方法来找到激进功能的界限。有些方法对激进功能起作用。使用直接替代是一种常见的方法。通过查找和取消分子和分母中的共同因素来改变不确定或未定义的形式,或者考虑和简化最大程度的变量能力,代表着共同的方法。

  播放段落

  One of the noteworthy differences between polynomial and radical functions is that the domain of polynomials can include all real values of the independent variable, but the domain of radical functions, e.g., $\begin{array}{r}\sqrt{x}\end{array}$ , is restricted.
  ::多元函数和激进函数之间的显著区别之一是,多元函数领域可以包括独立变量的所有实际价值,但激进函数领域,例如x,受到限制。

  播放段落

  Example 2
  ::例2

  播放段落

  Find $\begin{array}{r}\underset{x\to 0}{lim}\frac{\sqrt{x+4}-2}{x}\end{array}$ .
  ::查找 limx=0x+4-2x。

  播放段落

  Using direct substitution to find the limit of the function results in the indeterminate form $\begin{array}{r}\frac{0}{0}\end{array}$ . To transform the radical expression to a better form, do the following:
  ::使用直接替代来找到函数的极限,结果为不确定的00。要将激进表达式转换为更好的表达式,请采取以下行动:

  播放段落

  $$\begin{array}{rlrl}\underset{x\to 0}{lim}\frac{\sqrt{x+4}-2}{x}& =\underset{x\to 0}{lim}\left(\frac{\sqrt{x+4}-2}{x}\cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\right)& & \dots \text{Rationalize the numerator: multiply by}\\ & =\underset{x\to 0}{lim}\left(\frac{x+4-4}{x\cdot \sqrt{x+4}+2}\right)& & \text{the conjugate of the numerator.}\\ & =\underset{x\to 0}{lim}\left(\frac{x}{x\cdot \sqrt{x+4}+2}\right)\\ & =\underset{x\to 0}{lim}\frac{1}{\sqrt{x+4}+2}\\ & =\frac{1}{4}\end{array}$$

  
  ::=xxxxxxxxx+4xx+4xx+2x+4+4+2). 将分子合理化:乘以=limxx=0(x+4-4xxxx+4+2)

  播放段落

  Therefore, $\begin{array}{r}\underset{x\to 0}{lim}\frac{\sqrt{x+4}-2}{x}=\frac{1}{4}\end{array}$ .
  ::因此, limx=0x+4-2x=14。

  播放段落

  Review
  ::回顾

  播放段落

  Find each of the following limits if they exist.
  ::如果存在下列限制,请查找其中的每一项限制。

  播放段落
  1. $\begin{array}{r}\underset{x\to 3}{lim}\sqrt{x}\end{array}$
     ::limx3x 立方厘米
  播放段落
  2. $\begin{array}{r}\underset{x\to 8}{lim}\sqrt{x-7}\end{array}$
     ::limx8x-7
  播放段落
  3. $\begin{array}{r}\underset{x\to 4}{lim}\frac{\sqrt{x}-2}{x-4}\end{array}$
     ::limx4x-2x-4 立方厘米
  播放段落
  4. $\begin{array}{r}\underset{x\to 1}{lim}\frac{\sqrt{x+3}-2}{x-1}\end{array}$
     ::limx1x+3-2x-1
  播放段落
  5. $\begin{array}{r}\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{\sqrt{1+\sqrt{x}}-1}\end{array}$
     ::limx=0+x1+x-1
  播放段落
  6. $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt{x^2-5x}-x\right)\end{array}$
     ::limx(x2- 5x- x)
  播放段落
  7. $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{x^6+3x^2+1}}{4x^3+3}\end{array}$
     ::limxx6+3x2+14x3+3
  播放段落
  8. $\begin{array}{r}\underset{x\to 0}{lim}\frac{\sqrt{x+5}-\sqrt{5}}{x}\end{array}$
     ::limx=0x+5-5-5x
  播放段落
  9. $\begin{array}{r}\underset{x\to 3}{lim}\left(\sqrt{x^2+4x}\right)\end{array}$
     ::limx%3(x2+4x)
  播放段落
  10. $\begin{array}{r}\underset{x\to -1}{lim}(x^2+2x+10{)}^{\frac{3}{2}}\end{array}$
      ::limx1(x2+2x+10)32
  播放段落
  11. $\begin{array}{r}\underset{x\to 4}{lim}\frac{\sqrt{x+5}-3}{x-4}\end{array}$
      ::limx4x+5-5-3x-4
  播放段落
  12. $\begin{array}{r}\underset{x\to 1}{lim}\left(\sqrt{2x^3+3x^2+7}\right)\end{array}$
      ::limx%1( 2x3+3x2+7)
  播放段落
  13. $\begin{array}{r}\underset{x\to 3}{lim}\left(\sqrt[3]{2x^2-10}\right)\end{array}$
      ::立方厘米3( 2x2-2- 103)
  播放段落
  14. $\begin{array}{r}\underset{x\to 7}{lim}\frac{5x}{\sqrt{x+2}}\end{array}$
      ::立方厘米×75xxxx+2
  播放段落
  15. $\begin{array}{r}\underset{x\to 0}{lim}\frac{7-\sqrt{x^2+49}}{x}\end{array}$
      ::limx07-x2+49x

  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 。