2.5 多边和合理功能的限度

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 多边和合理功能的限度

  Collapse Expand

  多边和合理功能的限度

  播放段落

  Finding the limit of a polynomial function is relatively easy. Why? Finding the limit of a rational function can also be relatively easy, except under a few particular conditions. When is finding the limit of a rational function difficult?
  ::找到一个多元函数的极限比较容易。 为什么? 找到一个理性函数的界限也可以比较容易, 除非在几个特定条件下。 当找到一个理性函数的界限困难的时候?

  播放段落

  Limits of Polynomial and Rational Functions
  ::多边和合理功能的限度

  播放段落

  Polynomial Functions
  ::多元多边函数

  播放段落

  Recall that a function  $\begin{align*}f(x)\end{align*}$ polynomial function if it satisfies:
  ::回顾函数f(x)球体函数,如果满足如下条件:

  播放段落

  $$\begin{align*}f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2x^2+a_1x+a_0\end{align*}$$
  :x) = anxn+an- 1xn- 1a2x2+a1x+a0

  播放段落

  for all  $\begin{align*}x\end{align*}$ , where $\begin{align*}n\end{align*}$ is a non-negative integer and  $\begin{align*}a_0, a_1,a_2, \ldots, a_n\end{align*}$ are constant coefficients.
  ::对于所有 x, 其中 n 是非负整数, 和 10,a1,a2,... 是一个恒定系数 。

  播放段落

  As a refresher, use the limit properties to find $\begin{align*}\lim_{x \to 20} (x^2-3x+4)\end{align*}$ , i.e., the limit as  $\begin{align*}x\end{align*}$ approaches a particular value.
  ::作为复习器, 使用限制属性查找 limx20( x2 - 3x+4) , 即当 x 接近特定值时, 限制值 。

  播放段落

  The function is a polynomial, a quadratic trinomial that is graphed below, and can be treated as the sum of three functions. This means that we can use the rule “the limit of the sum is the sum of the limits” in the determination of the limit.
  ::函数是一个多面函数,一个四边三角函数,如下文所示,可以作为三个函数的总和处理。这意味着我们可以在确定限制时使用规则“总和的限度是限度的和”。

  

  播放段落

  Since, the polynomial can be treated as the sum of three functions, we can use the property “the limit of the sum is the sum of the limits” in the determination of the limit.
  ::由于多功能制可被视为三功能之和,我们可以在确定限额时使用该财产“总金额的限额是限额之和”。

  播放段落

  $$\begin{align*}\lim_{x \to 20} (x^2-3x+4) & = \lim_{x \to 20} x^2 + \lim_{x \to 20} (-3x) + \lim_{x \to 20} 4\\ & = ( \lim_{x \to 20} x )^2 -3 \lim_{x \to 20} x+4\\ & = (20)^2 - 3(20)+4\\ & = 344\end{align*}$$
  :x2 - 3x+4) = limx20x2+limx2020(-3x) +limxXX20(-3x) +limx= 204= (limx20x) 2 - 3limx20x+4= (2020) 2 - 3x+4= (20) 3x+4= 344

  播放段落

  Therefore, $\begin{align*}\lim_{x \to 20} (x^2-3x+4) = 344\end{align*}$ .
  ::因此,limx%20(x2-3x+4)=344。

  播放段落

  Note that the value of this limit could have been found by direct substitution of  $\begin{align*}x=1\end{align*}$ in the polynomial function. 
  ::请注意,这一限值的值可以通过在多面函数中直接替换 x=1 来找到。

  播放段落

  Now, find  $\begin{align*}\lim_{x \to \infty} (x^2-3x+4)\end{align*}$ , i.e., the limit as  $\begin{align*}x\end{align*}$ approaches infinity. This is looking at end behavior.
  ::现在, 找到 limx( x2 - 3x+ 4) , 即 xoproaches Infinity 的限值 。 它正在查看最终行为 。

  播放段落

  The polynomial can be treated as the product of two functions. This means that we can use the rule “the limit of the product of functions is the product of the limits of each function” in the determination of the limit.
  ::多功能可被视为两个功能的产物,这意味着我们可以在确定限制时使用“功能产物的极限是每种功能的极限的产物”的规则。

  播放段落

  $$\begin{align*}\lim_{x \to \infty} (x^2-3x+4) & = \lim_{x \to \infty} x^2 \left (1+ \frac{-3x}{x^2} + \frac{4}{x^2} \right )\\ & = \lim_{x \to \infty} x^2 \cdot \lim_{x \to \infty} \left (1+ \frac{-3x}{x^2} + \frac{4}{x^2} \right )\\ & = \infty \cdot 1 \qquad \qquad \ldots \text{Note that the terms divided by }x^2 \text{ go to 0}.\\ & = \infty\end{align*}$$
  ::limx(x2-3x+4) =limxx2(1) 13x2+4x2) =limxxx22xlimx(13xx2+4x2) 1...注意条件除以 x2 = 0。

  播放段落

  Therefore, $\begin{align*}\lim_{x \to \infty} (x^2-3x+4) = \infty\end{align*}$ . A similar evaluation shows that $\begin{align*}\lim_{x \to -\infty} (x^2-3x+4) = \infty\end{align*}$ . This behavior depicts the fact that the end behavior of polynomials goes as the term with the highest degree, and values grow without bound.
  ::因此, limx( x2 - 3x+4)\\\\\\\\\\\\\\\\\\\ x2 - 3x+4)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x2 - 3x+4\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

  播放段落

  The results of these two examples can be generalized to the following properties:
  ::这两个例子的结果可概括如下:

  播放段落

  Given the polynomial function $\begin{align*}f(x) =p(x)\end{align*}$ :
  ::给定多元函数 f( x) = p( x) :

  播放段落
  1. $\begin{align*}\lim_{x \to a} f(x) = p(a)\end{align*}$ , for any real number $\begin{align*}a\end{align*}$ .
     ::=p(a),任何实际数字a。
  播放段落
  2. $\begin{align*}\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} a_nx^n \ \infty \text{ or } -\infty\end{align*}$ .
     ::立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺 = 立方公尺

  播放段落

  Rational Functions
  ::理性函数

  播放段落

  Now let’s consider limits of rational functions. A rational function is the ratio of two polynomials. In the case of a single variable, $\begin{align*}x\end{align*}$ , a function is called a rational function if and only if it can be written in the form:
  ::现在让我们来考虑理性函数的限度。 理性函数是两个多元函数的比。 在单一变量 x 的情况下,函数被称为理性函数,如果并且只有在能够以形式写成时:

  播放段落

  $$\begin{align*}f(x) = \frac{P(x)}{Q(x)} = \frac{a_nx^n+a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1} + \cdots + b_2x^2 + b_1x+b_0}\end{align*}$$
  ::f( x) = P( x) Q( x) = anxn+an- 1xn- 1xn- 1xn- 1a2x2+a1x+a0bmxm+bm- 1xm- 1xm+bm- 1xm- 1b2x2+b1x+b0

  播放段落

  where  $\begin{align*}P(x)\end{align*}$ and $\begin{align*}Q(x)\end{align*}$ are polynomial functions in  $\begin{align*}x\end{align*}$ and  $\begin{align*}Q(x)\end{align*}$ is non-zero. The domain of  $\begin{align*}f\end{align*}$ is the set of all values of  $\begin{align*}x\end{align*}$ for which the denominator  $\begin{align*}Q(x)\end{align*}$ is not zero.
  ::P(x) 和 Q(x) 是 x 中的多元函数, Q(x) 非零 。 f 的域是 x 的一组全部值, 其分母 Q(x) 不是零 。

  播放段落

  Find $\begin{align*}\lim_{x \to 10} \frac{x-4}{x-2}\end{align*}$ .
  ::查找 limx10x-4x-2。

  播放段落

  The function is shown in the graph below.
  ::函数在下图中显示。

  

  播放段落

  We note first that the denominator of the rational function is not zero at the value $\begin{align*}x=10\end{align*}$ . The quotient rule can therefore be used to start the evaluation of the function as follows:
  ::我们首先注意到,理性函数的分母在值 x=10 时不是零。 因此,可以使用商数规则开始对函数进行如下评价:

  播放段落

  $$\begin{align*}\lim_{x \to 10} \frac{x-4}{x-2} & = \frac{\lim\limits_{x \to 10} (x-4)}{\lim\limits_{x \to 10} (x-2)}\\ & = \frac{\lim\limits_{x \to 10} (x) - \lim\limits_{x \to 10} (4)}{\lim\limits_{x \to 10} (x) - \lim\limits_{x \to 10} (2)}\\ & = \frac{(10-4)}{(10-2)}\\ & = \frac{3}{4}\end{align*}$$
  :x-4)limx10(x-2)10(x-2)=limx10(x)-limx10(x)-limx10(x)-10(limx)10(x)-limx10(x)-limx10(x)-(10)-(10)-(10)-(10)-(10)-(2)=(10-4(10-2)=34)

  播放段落

  Therefore $\begin{align*}\lim_{x \to 10} \frac{x-4}{x-2} = \frac{3}{4}\end{align*}$ .
  ::因此,10x-4x-2=34。

  播放段落

  Note that because the denominator does not equal 0 at $\begin{align*}x=10\end{align*}$ , the limit could have been found by direct substitution of  $\begin{align*}x=10\end{align*}$ in the rational function.
  ::请注意,由于分母在 x=10 时不等于 0,因此可以通过在合理函数中直接替换 x=10 来找到这一限值。

  播放段落

  Now, find the end behavior of that same function, i.e. find  $\begin{align*}\lim_{x \to \infty} \frac{x-4}{x-2}\end{align*}$ .
  ::现在,找到相同函数的结束行为, 即找到 limxx- x- 4x-2 。

  播放段落

  The following steps are used to evaluate the limit at as  $\begin{align*}x\end{align*}$ approaches infinity.
  ::以下步骤用于评估x方针的无限限值。

  播放段落

  $$\begin{align*}\lim_{x \to \infty} \frac{x-4}{x-2} & = \lim\limits_{x \to \infty} \frac{x \left ( 1-\frac{4}{x} \right ) }{x \left ( 1-\frac{2}{x} \right )} \quad \ldots \text{Factor the out the highest degree variable}\\ & \qquad \qquad \qquad \qquad \qquad \text{from numerator and denominator.}\\ & = \frac{\lim\limits_{x \to \infty} \left ( 1-\frac{4}{x} \right )}{\lim\limits_{x \to \infty} \left ( 1-\frac{2}{x} \right )} \quad \ \ \ldots \text{Use the quotient rule of limits.}\\ & = \frac{1-0}{1-0}\\ & = 1 \end{align*}$$
  ::limxxx-4x-2=limxxxxxx(1-4x)x(1-2-2x)x...从分子和分母中显示最高度变量。=limx{(1-4x)limx(1-2-2x)...使用数数限制规则。=1-01-0=1

  播放段落

  Therefore $\begin{align*}\lim_{x \to \infty} \frac{x-4}{x-2}=1\end{align*}$ .
  ::因此, limxx- 4x-2= 1。

  播放段落

  The above problems illustrate the evaluation of the limit of a rational function at a value of  $\begin{align*}x\end{align*}$ for which the denominator does not equal 0. Sometimes finding the limit of a rational function  $\begin{align*}f(x)\end{align*}$ at some  $\begin{align*}x=a \end{align*}$ can entail more work than just direct substitution because the denominator equals zero at $\begin{align*}x=a \end{align*}$ . What if the denominator is equal to 0?
  ::上述问题说明了对x值合理函数极限值的评价,对于x值,分母并不等于0。有时在某种 x=a 找到合理函数f(x)f(x)的极限,可能比直接替代需要更多工作,因为分母在 x=a 等于0。如果分母等于0,则该分母等于0?

  播放段落

  Take the problem:   $\begin{align*}\lim_{x \to 2} \frac{x^2-4}{x-2}\end{align*}$ .
  ::问题在于: limx% 2x2- 4x-2。

  播放段落

  Notice that the function here is indeterminate at $\begin{align*}x = 2\end{align*}$ , so that direct substitution does not work. However, in this case it is possible to remove the zero in the denominator by factoring the numerator and canceling the factor  $\begin{align*}(x - 2)\end{align*}$ from both the numerator and the denominator.
  ::请注意此函数在 x=2 时未确定, 直接替换无效 。 但是, 在此情况下, 可以通过乘数计算器来去除分母中的零, 并从分子和分母中取消系数( x-2 ) 。

  播放段落

  $$\begin{align*}\lim_{x \to 2} \frac{x^2-4}{x-2} & = \lim_{x \to 2} \frac{(x-2)(x+2)}{(x-2)} \quad \ldots \text{Factor the numerator.}\\ & = \lim_{x \to 2} (x+2)\\ & = 4\end{align*}$$
  ::limx2x2-4x-2=limx22(x-2)(x+2)(x)(x-2)(x-2).。分子的分数。=limx=2(x+2)=4

  播放段落

  Therefore, $\begin{align*}\lim_{x \to 2} \frac{x^2-4}{x-2}=4\end{align*}$ .
  ::因此, limx%2x2-4x-2=4。

  播放段落

  The factoring of the numerator shown above, and then cancelling any common factors in the denominator, is a common technique used to find the limits of rational functions at points where the denominator is 0. Always check to see if the function can be simplified to remove the zero in the denominator, especially by cancelling common a factor that removes a discontinuity.
  ::上面显示的分子系数,然后取消分母中的任何共同系数,是一种常见技术,用于在分母为0的点找到理性函数的极限。总是检查该函数是否可以简化,以去除分母中的零,特别是通过取消一个消除不连续性的共同系数。

  播放段落

  Now, find   $\begin{align*}\lim_{x \to \infty} \frac{x^2-4}{x-2}\end{align*}$  .  We look at the end behavior of the function as  $\begin{align*}x\end{align*}$ goes to infinity as follows:
  ::现在, 找到 limxx2 - 4x-2 。 我们从 x 到无穷度时查看函数的结束行为 如下 :

  播放段落

  $$\begin{align*}\lim_{x \to \infty} \frac{x^2-4}{x-2} & = \lim_{x \to \infty} \frac{x^2\left ( 1-\frac{4}{x^2} \right )}{x \left ( 1-\frac{2}{x} \right )} \qquad \quad \ldots \text{Factor the highest degree variable}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \text{in numerator and denominator.}\\ & = \lim_{x \to \infty} \frac{x\left ( 1-\frac{4}{x^2} \right )}{ \left ( 1-\frac{2}{x} \right )}\\ & = \lim_{x \to \infty} x \cdot \lim_{x \to \infty} \frac{\left ( 1-\frac{4}{x^2} \right )}{ \left ( 1-\frac{2}{x} \right )} \quad \ldots \text{Use limit of product rule}\\ & = \infty \cdot 1\\ & = \infty\end{align*}$$
  ::=limxxxxx2-4x-2=limxxxxxx2(1-4x2)xx(1-2-2xx)...最高度变量数和分母。=limxxxxx(1-4x2)(1-2x)=limxxxxxxxxxxxxxxxxlxxílxxxxxxxxx2(1-4x2)x(1-2x)...产品规则的使用限值*1_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

  播放段落

  Therefore, $\begin{align*}\lim_{x \to \infty} \frac{x^2-4}{x-2} = \infty\end{align*}$ .
  ::因此,Limxxx2 -4x-2。

  播放段落

  The end behavior as  $\begin{align*}x\end{align*}$ goes to $\begin{align*}-\infty\end{align*}$  can be determined by a similar approach, and is found to be $\begin{align*}-\infty\end{align*}$ .
  ::x 到 * 的结束行为可以通过类似的方法来决定, 并被发现是 \ 。

  播放段落

  Definition of the Limit of Rational Functions
  ::合理职能限制的定义

  播放段落

  For the rational function $\begin{align*}f(x) = \frac{p(x)}{q(x)}\end{align*}$  and any real number $\begin{align*}a\end{align*}$ ,
  ::对于合理的函数 f( x) = p( x)q( x) 和任何实际数字 a,

  播放段落

  $$\begin{align*}\lim_{x \to a} f(x) = \frac{p(a)}{q(a)} \text{ if } q(a) \ne 0.\end{align*}$$
  :x) = p(a)q(a) if q(a) =0。

  播放段落

  If $\begin{align*}q(a)=0\end{align*}$ , then the function may or may not have a limit.
  ::如果 q(a) =0,则该函数可能有或没有限制。

  播放段落

  For the rational function $\begin{align*}f(x) = \frac{p(x)}{q(x)} = \frac{a_nx^n+a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1} + \cdots + b_2x^2 + b_1x+b_0}\end{align*}$ ,
  ::对于合理的函数 f( x) = p( x) q( x) = anxn+an- 1xn - 1a2x2+a1x+a1x+a0bmxm+bm- 1xm- 1b2x2+b1x+b0,

  播放段落

  $$\begin{align*}\lim_{x \to \pm \infty } f(x) = \left ( \frac{a_n}{b_m} \right ) \cdot \lim_{x \to \infty} x^{n-m} = \begin{cases} + \infty \text{ or } - \infty & \text{if } n>m \\ \frac{a_n}{b_m} & \text{if } n=m \\ 0 & \text{if } n < m . \end{cases}\end{align*}$$
  ::limxf(x) =( anbm) = limxxn- m 或 @ if n> manbmif n= m0if nm 。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked about the respective difficulties of finding the limit of polynomial and rational functions. Finding the limit of a polynomial function is relatively easy because a polynomial function can be evaluated at any value of the independent variable so that the limit at a specific value can be evaluated by direct substitution. The limit as the independent variable goes to  $\begin{align*}\pm \infty\end{align*}$ is just  $\begin{align*}\pm \infty\end{align*}$ depending whether the degree of the polynomial is even or odd.
  ::早些时候,有人询问了找到多元和理性函数的极限的各自困难。 找到多元函数的界限相对容易, 因为可以按照独立变量的任何值来评估多元函数, 从而可以通过直接替代来评估特定值的界限。 独立变量进入 {} 的极限只是 {} 取决于多元函数的大小是偶数还是奇数 。

  播放段落

  Evaluating the limit of a rational function can be more difficult because direct substitution may lead to an undefined or indeterminate form that requires a different approach, and the limit as the independent variable goes to $\begin{align*}\pm \infty\end{align*}$  depends on which is larger, the degree of the numerator polynomial or the degree of denominator polynomial.
  ::评估合理功能的限度可能更加困难,因为直接替代可能导致一种不确定或不确定的形式,需要不同的方法,而独立变量去到的限度取决于哪一个更大,即分子多位数的程度或分母多位数的程度。

  播放段落

  Example 2
  ::例2

  播放段落

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

  播放段落

  The numerator and the denominator are both equal to zero at $\begin{align*}x = 3\end{align*}$ , but there is a common factor  $\begin{align*}x - 3\end{align*}$ 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):

  播放段落

  $$\begin{align*}\lim_{x \to 3} \frac{2x-6}{x^2+x-12} & = \lim_{x \to 3} \frac{2(x-3)}{(x+4)(x-3)}\\ & = \lim_{x \to 3} \frac{2}{x+4}\\ & = \frac{2}{7}\end{align*}$$
  ::立方公尺xxxxx3x-3(x-3)(x-4)(x-3)(x-3)=limx=32x+4=27

  播放段落

  Example 3
  ::例3

  播放段落

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

  播放段落

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

  播放段落

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

  播放段落

  Since we have  $\begin{align*}(x - 1)\end{align*}$ 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(x-1)都有(x-1),我们知道原始功能等于仅仅-5x-4,除非未定义(1)。

  播放段落

  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。

  播放段落

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

  播放段落

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

  播放段落

  Example 4
  ::例4

  播放段落

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

  播放段落

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

  播放段落

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

  播放段落

  Since we have  $\begin{align*}(x + 2)\end{align*}$  in both numerator and denominator, we know that the original function is equal to just  $\begin{align*}-x-4\end{align*}$
  ::由于我们在分子和分分母中都有(x+2)分数,我们知道原始函数等于-x-4

  播放段落

  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。

  播放段落

  To find the value, just solve  $\begin{align*}-x-4\end{align*}$  for  $\begin{align*}x=-2\end{align*}$
  ::要找到值, 只需为 x @% 2 解答 ~ x_ 4

  播放段落

  $\begin{align*}\therefore \lim_{x \to -2} \frac{-x^2+2x+8}{x+2} = -(\text{-2}) - 4 \to \text{-2}\end{align*}$
  ::2 -x2+2x+8x+2(-2) -42

  播放段落

  Review
  ::回顾

  播放段落

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

  播放段落
  1. $\begin{align*}\lim_{x \to 1} \frac{-12x^2+12}{4x-4}\end{align*}$
     ::limx=1 - 12x2+124x- 4
  播放段落
  2. $\begin{align*}\lim_{x \to 2} \frac{\frac{3}{x+3} - \frac{11}{9}}{2x-4}\end{align*}$
     ::立方厘米23x+3-1192x-4
  播放段落
  3. $\begin{align*}\lim_{x \to \frac{-57}{56}} \frac{\frac{-5x-3}{2x+3}- \frac{13}{6}}{-56x-57}\end{align*}$
     ::5756-55x-32x+3-136-56x-57
  播放段落
  4. $\begin{align*}\lim_{x \to 2} \frac{2x^2-5x+2}{-x+2}\end{align*}$
     ::limx% 22x2 - 5x+2 - x+2
  播放段落
  5. $\begin{align*}\lim_{x \to \frac{3}{4}} \frac{4x^2+5x-6}{4x-3}\end{align*}$
     ::立方厘米x344x2+5x-64x-3
  播放段落
  6. $\begin{align*}\lim_{x \to 4} \frac{\frac{-3}{-2x+3}-\frac{3}{5}}{6x-24}\end{align*}$
     ::立方厘米4 - 3 - 2x+3 - 356x-24
  播放段落
  7. $\begin{align*}\lim_{x \to \frac{-3}{2}} \frac{\frac{-4x-3}{-2x+2} - \frac{5}{2}}{-2x-3}\end{align*}$
     ::32 - 4x - 3 - 2x+2 - 52 - 2x - 3
  播放段落
  8. $\begin{align*}\lim_{x \to 4} \frac{x^2-8x+16}{x-4}\end{align*}$
     ::limx=4x2-8x+16x-4
  播放段落
  9. $\begin{align*}\lim_{x \to \frac{10}{39}} \frac{\frac{3x+3}{-3x+4} - \frac{7}{6}}{39x-10}\end{align*}$
     ::10393x+3-3-3x+4-7639x-10
  播放段落
  10. $\begin{align*}\lim_{x \to -4} \frac{-3x^2+7x-20}{-x-4}\end{align*}$
      ::4 - 3x2+7x - 20 - x- 4
  播放段落
  11. $\begin{align*}\lim_{x \to -4} \frac{4x^2+14x-8}{x+4}\end{align*}$
      ::limx44x2+14x-8x+4
  播放段落
  12. $\begin{align*}\lim_{x \to \frac{-18}{13}} \frac{\frac{-4x+1}{-3x+5} - \frac{5}{7}}{-13x-18}\end{align*}$
      ::1813-4x+1-3x+5-5-57-57-13x-18
  播放段落
  13. $\begin{align*}\lim_{x \to -2} \frac{\frac{3}{x+4} - \frac{3}{2}}{-3x-6}\end{align*}$
      ::立方 23x+4-32-3x-6
  播放段落
  14. $\begin{align*}\lim_{x \to \frac{-3}{4}} \frac{\frac{-x+3}{4x+3}- \frac{11}{2}}{-4x-3}\end{align*}$
      ::立方公尺xxxxx34x+3-112-4x-3
  播放段落
  15. $\begin{align*}\lim_{x \to \frac{1}{4}} \frac{-8x^2-2x+1}{-4x+1}\end{align*}$
      ::立方公尺x14-8x2-2x+1-4x+1
  播放段落
  16. $\begin{align*}\lim_{x \to \frac{1}{4}} \frac{16x^2-16x+3}{-4x+1}\end{align*}$
      ::立方厘米1416x2 - 16x+3 - 4x+1
  播放段落
  17. $\begin{align*}\lim_{x \to 0} \frac{x^2+3x}{x}\end{align*}$
      ::limx=0x2+3xxx

  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 。