课程： 8.5 合理函数限制

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逻辑函数限制

Finding the limit of a rational function is actually much less complex than it may seem, in fact many of the limits you have already evaluated have been rational functions.
::找到理性功能的极限实际上比看起来要复杂得多,事实上,你已经评估的许多界限是理性功能。

In this lesson, you will gain more experience working with rational function limits, and will use another theorem which simplifies the process of finding the limit of some rational functions.
::在这个教训中,你将获得更多关于合理功能限制的经验,并将使用另一个理论来简化寻找某种合理功能限制的过程。

Rational Function Limits
::逻辑函数限制

Sometimes finding the limit of a rational function at a point a is difficult because evaluating the function at the point a leads to a denominator equal to zero. The box below describes finding the limit of a rational function.
::有时很难在某个点上找到合理函数的极限,因为对点上的函数进行评价会导致分母等于零。下面的框说明找到合理函数的极限。

Theorem: The Limit of a Rational Function For the rational function $\begin{align}f(x) = \frac{p(x)} {q(x)}\end{align}$ and any real number a , $\begin{align}\lim_{x \rightarrow a} f(x) = \frac{p(a)} {q(a)} \text{ if } q(a) \neq 0\end{align}$ . However, if $\begin{align} \,\! q(a) = 0 \end{align}$ then the function may or may not have any outputs that exist.

Examples
::实例

Example 1
::例1

Find $\begin{align*}\lim_{x \rightarrow 3} \frac{2 - x} {x - 2}\end{align*}$ .
::寻找 limx32 -xx-2 。

Using the theorem above, we simply substitute x = 3: $\begin{align*}\lim_{x \rightarrow 3} \frac{2 - x} {x - 2} = \frac{2 - 3} {3 - 2} = -1\end{align*}$
::使用上面的定理, 我们只需替换 x = 3: limx- 32-xx-2= 2- 33-21

Example 2
::例2

Find $\begin{align*}\lim_{x \rightarrow 3} \frac{x + 1} {x - 3}\end{align*}$ .
::查找 limx%3x+1x- 3 。

Notice that the domain of the function is continuous (defined) at all real numbers except at x = 3. If we check the we see that $\begin{align*}\lim_{x \rightarrow 3^{+}} \frac{x + 1} {x - 3}=\infty\end{align*}$ and $\begin{align*}\lim_{x \rightarrow 3^{-}} \frac{x + 1} {x - 3}=-\infty\end{align*}$ . Because the one-sided limits are not equal, the limit does not exist .
::请注意,函数的域除x = 3. 外,所有实际数字都是连续的(定义的),如果我们检查我们是否看到 limx3+x+1x-3 和 limx3-x+1x-3 和 limx3-x+1x-3 。由于单向限制不相等,因此该限制并不存在。

Example 3
::例3

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

Notice that the function here is discontinuous at x = 2, that is, the denominator is zero at x = 2. However, it is possible to remove this discontinuity by canceling the factor x - 2 from both the numerator and the denominator and then taking the limit:
::注意此函数在 x = 2 时不连续, 也就是说, 分母在 x = 2 时为 0 = 2. 但是, 可以通过从分子和分母中取消乘数 x - 2 来消除这种不连续性, 然后选择限制 :

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

This is a common technique used to find the limits of rational functions that are discontinuous at some points. When finding the limit of a rational function, always check to see if the function can be simplified.
::这是一种常见的方法,用来找到理性功能的极限,而理性功能在某些时点是不连续的。当找到理性功能的极限时,总是要检查是否可以简化功能。

Example 4
::例4

Find $\begin{align*}\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12}\end{align*}$ .
::查找 limx% 32x- 6x2+x- 12 。

The numerator and the denominator are both equal to zero at x = 3, but there is a common factor x - 3 that can be removed (that is, we can simplify the rational function):
::分子和分母在 x = 3 时均等于零,但有一个共同系数 x - 3 可以删除(即我们可以简化理性功能):

	$\begin{align}\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12}\end{align}$	$\begin{align}= \lim_{x \rightarrow 3} \frac{2(x - 3)} {(x + 4) (x - 3)}\end{align}$
		$\begin{align}= \lim_{x \rightarrow 3} \frac{2} {x + 4}\end{align}$
		$\begin{align}= \frac{2} {7}\end{align}$

Example 5
::例5

Find $\begin{align*}\lim_{x \to 1} \frac{-5x^2 +x +4}{x - 1}\end{align*}$ .
::查找 limx% 1 - 5x2+x+4x- 1 。

$\begin{align*}\frac{(-5x - 4)(x - 1)}{(x - 1)}\end{align*}$ ..... Start by factoring the numerator
:-5x-4) (x-1) (x- 1) (x- 1) ....。从乘数开始

Since we have (x - 1) in both numerator and denominator, we know that the original function is equal to just $\begin{align*}-5x -4\end{align*}$ except where it is undefined (1).
::由于我们在分子和分母两方面都有(x-1),我们知道原始功能等于仅仅-5x-4,除非未定义(1)。

Therefore the closer we get to inputing 1, the closer we get to the same value, whether from the + or - side.
::因此,我们越接近输入1, 我们就越接近相同的价值, 无论是从+或-侧。

To find the value, just solve $\begin{align*}-5x - 4\end{align*}$ for $\begin{align*}x = 1\end{align*}$ .
::要找到值, 只需解析 x=1 的 5x- 4 。

$\begin{align*}\therefore \lim_{x \to 1} \frac{-5x^2 +x +4}{x - 1} = -5 \cdot 1 - 4 \to -9\end{align*}$
::*%x%1 - 5x2+x+4x - 1*5*1 - 4*9

Example 6
::例6

Find $\begin{align*}\lim_{x \to -2} \frac{-x^2 + 2x + 8}{x + 2}\end{align*}$ .
::查找limx% 2 - x2+2x+8x+2。

$\begin{align*}\frac{(-x + 4)(x + 2)}{(x + 2)}\end{align*}$ ..... Start by factoring the numerator
:- x+4) (x+2) (x+2) (x+2) ....。从乘数开始

Since we have (x + 2) in both numerator and denominator, we know that the original function is equal to just $\begin{align*}-x + 4\end{align*}$ except where it is undefined (-2).
::由于我们在分子和分母两方面都有(x+2),我们知道原始函数等于只-x+4,除非未定义(-2)。

Therefore the closer we get to substituting -2, the closer we get to the same output value, whether from the + or - side.
::因此,我们越接近于取代 -2, 我们就越接近于相同的产出值, 无论是从+ 或- 侧。

To find the value, just solve $\begin{align*}-x + 4\end{align*}$ for $\begin{align*}x = -2\end{align*}$ .
::要找到值, 只需解析 x% 2 的 ~ x+4 。

$\begin{align*}\therefore \lim_{x \to -2} \frac{-x^2 + 2x + 8}{x + 2} = -(-2) + 4 \to 6\end{align*}$
::2 -x2+2x+8x+2(-2)+46

Review
::回顾

Solve the following rational function limits.
::解决以下合理功能限制。

$\begin{align*}\lim_{x \to 1} \frac{-12x^2 + 12}{4x - 4}\end{align*}$
::limx=1 - 12x2+124x- 4

$\begin{align*}\lim_{x \to 2} \frac{\frac{3}{x+3} - \frac{11}{9}}{2x - 4}\end{align*}$
::立方厘米23x+3-1192x-4

$\begin{align*}\lim_{x \to \frac{-57}{56}} \frac{\frac{-5x - 3}{2x + 3} - \frac{13}{6}}{-56x - 57}\end{align*}$
::5756-55x-32x+3-136-56x-57

$\begin{align*}\lim_{x \to 2} \frac{2x^2 - 5x + 2}{-x + 2}\end{align*}$
::limx% 22x2 - 5x+2 - x+2

$\begin{align*}\lim_{x \to \frac{3}{4}} \frac{4x^2 + 5x - 6}{4x - 3}\end{align*}$
::立方厘米x344x2+5x-64x-3

$\begin{align*}\lim_{x \to 4} \frac{\frac{-3}{-2x + 3} - \frac{3}{5}}{6x - 24}\end{align*}$
::立方厘米4 - 3 - 2x+3 - 356x-24

$\begin{align*}\lim_{x \to \frac{-3}{2}} \frac{\frac{-4x - 3}{-2x+2} - \frac{5}{2}} {-2x -3}\end{align*}$
::32 - 4x - 3 - 2x+2 - 52 - 2x - 3

$\begin{align*}\lim_{x \to 4} \frac{x^2 - 8x + 16}{x - 4}\end{align*}$
::limx=4x2-8x+16x-4

$\begin{align*}\lim_{x \to\frac{10}{39} } \frac{\frac{3x + 3}{-3x + 4} - \frac{7}{6}} {39x - 10}\end{align*}$
::10393x+3-3-3x+4-7639x-10

$\begin{align*}\lim_{x \to -4} \frac{3x^2 + 7x - 20}{-x - 4}\end{align*}$
::43x2+7x-20-x-4

$\begin{align*}\lim_{x \to -4} \frac{4x^2 + 14x -8}{x + 4}\end{align*}$
::limx44x2+14x-8x+4

$\begin{align*}\lim_{x \to \frac{-18}{13}} \frac{\frac{-4x + 1}{-3x + 5} - \frac{5}{7}} {-13x - 18}\end{align*}$
::1813-4x+1-3x+5-5-57-57-13x-18

$\begin{align*}\lim_{x \to -2} \frac{\frac{3}{x + 4} - \frac{3}{2}}{-3x - 6}\end{align*}$
::立方 23x+4-32-3x-6

$\begin{align*}\lim_{x \to \frac{-3}{4}} \frac{\frac{-x + 3}{4x + 3} - \frac{11}{2}}{-4x - 3}\end{align*}$
::立方公尺xxxxx34x+3-112-4x-3

$\begin{align*}\lim_{x \to \frac{1}{4}} \frac{-8x^2 - 2x + 1}{-4x + 1}\end{align*}$
::立方公尺x14-8x2-2x+1-4x+1

$\begin{align*}\lim_{x \to \frac{1}{4}} \frac{16x^2 - 16x + 3}{-4x + 1}\end{align*}$
::立方厘米1416x2 - 16x+3 - 4x+1

$\begin{align*}\lim_{x \to 0} \frac{x^2 + 3x}{x}\end{align*}$
::limx=0x2+3xxx

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。

	$\begin{align}\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12}\end{align}$	$\begin{align}= \lim_{x \rightarrow 3} \frac{2(x - 3)} {(x + 4) (x - 3)}\end{align}$
		$\begin{align}= \lim_{x \rightarrow 3} \frac{2} {x + 4}\end{align}$
		$\begin{align}= \frac{2} {7}\end{align}$

章节大纲

常规

逻辑函数限制

Rational Function Limits ::逻辑函数限制

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Review ::回顾

Review (Answers) ::回顾(答复)

Resources ::资源