2.7 涉及三角函数的限制

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  涉及三角函数的限制

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  Trigonometric functions can be a component of an expression and therefore subject to a limit process. Do you think that the periodic nature of these functions, and the limited or infinity range of individual trigonometric functions would make evaluating limits involving these functions difficult?
  ::三角函数可以是表达式的组成部分,因此受一定过程的限制。 您是否认为这些函数的周期性以及个别三角函数的有限或无限范围会使得评估这些功能的局限性变得困难?

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  Limits with Trigonometric Functions
  ::带有三角函数限制的三角函数

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  The limit rules presented in earlier concepts offer some, but not all, of the tools for evaluating limits involving trigonometric functions.
  ::先前概念中提出的限制规则提供了一些但并非全部的工具,用以评价涉及三角函数的限制。

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  Let's find  the following  limits:
  ::让我们找出以下限制:

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  1. $\begin{array}{r}\underset{x\to 0}{lim}\sin (x)\end{array}$
     ::立方公尺 [x]
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  2. $\begin{array}{r}\underset{x\to 0}{lim}\cos (x)\end{array}$
     ::立方公尺 [x]
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  3. $\begin{array}{r}\underset{x\to \pm \mathrm{\infty }}{lim}\sin (x)\end{array}$
     ::limxsin(x)

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  We can find these limits by evaluating the function as  $\begin{array}{r}x\end{array}$ approaches 0 on the left and the right, i.e., by evaluating the two . The graphs and tables values are shown below.
  ::通过对左边和右边的 x 接近0 的函数进行评估,即对这两个函数进行评估,我们可以找到这些限制。以下显示图表和表格的数值。

  $$\begin{array}{r}x(rad)\end{array}$$	-0.001	-0.0001	0	0.0001	0.001
  播放段落 $\begin{array}{r}\sin (x)\end{array}$ :x) 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别	-0.001	-0.0001	0	0.0001	0.001
  播放段落 $\begin{array}{r}\cos (x)\end{array}$ ::COs(x)	0.999	0.9999	1	0.9999	0.999

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  Inspection of the graph below, and table of values in the vicinity of $\begin{array}{r}x=0\end{array}$ , indicates that:
  ::对下图的检查和x=0附近的数值表表明:

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  1. $\begin{array}{r}\underset{x\to 0^{-}}{lim}\sin (x)=\underset{x\to 0^{+}}{lim}\sin (x)=0\end{array}$ , which means $\begin{array}{r}\underset{x\to 0}{lim}\sin (x)=0\end{array}$ .
     ::=0=0 =0 = limx=0 =0 = = limx=0 = (x)=0 = limx=0 = (x)=0。
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  2. $\begin{array}{r}\underset{x\to 0^{-}}{lim}\cos (x)=\underset{x\to 0^{+}}{lim}\cos (x)=1\end{array}$ , which means $\begin{array}{r}\underset{x\to 0}{lim}\cos (x)=1\end{array}$ .
     ::立方公尺= 立方公尺= 立方公尺= 立方公尺= 立方尺= 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

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  Note that the limits can be found using direct substitution.
  ::请注意,使用直接替代可以找到这些限度。

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  3. Because  $\begin{array}{r}\sin (x)\end{array}$ is a periodic function, as  $\begin{array}{r}x\end{array}$ gets larger (smaller) and larger (smaller), its value will oscillate between 1 and -1, and never settle to a single value. Therefore, we can say that $\begin{array}{r}\underset{x\to \pm \mathrm{\infty }}{lim}\sin (x)\end{array}$   does not exist .
     ::因为sin(x) 是一个周期函数, 因为x越大( 越小) 越大( 越小) , 其价值在 1 和 - 1 之间摇摆不定, 并且永远不能满足于单一值 。 因此, 我们可以说, limx sin(x) 不存在 。

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  We can generalize and extend the findings above and present the following properties:
  ::我们可以概括和扩展上述调查结果,并提出以下属性:

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  Limit Properties for Basic Trigonometric Functions
  ::基本三角函数的限制属性

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  • Limit as $\begin{array}{r}x\to a\end{array}$  for any real  $\begin{array}{r}a\end{array}$ :
    ::任何真实 a 的 xa 限制为 :

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  $$\begin{array}{rl}& \underset{x\to a}{lim}\sin (x)=\sin (a)\underset{x\to a}{lim}\cos (x)=\cos (a)\\ & \underset{x\to a}{lim}\tan (x)=\tan (a)\underset{x\to a}{lim}\cot (x)=\cot (a)\\ & \underset{x\to a}{lim}\sec (x)=\sec (a)\underset{x\to a}{lim}\csc (x)=\csc (a)\end{array}$$

  
  :x)=sin(a) (x) (a) (a) (a) (a) (x) (a) (a) (x) (a) (a) (x) (a) (a) (x) (csc) (a)

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  • Limit as $\begin{array}{r}x\to \pm \mathrm{\infty }\end{array}$ :
    ::限制为 x {} :

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  $$\begin{array}{r}\underset{x\to \pm \mathrm{\infty }}{lim}\sin (x)\underset{x\to \pm \mathrm{\infty }}{lim}\cos (x)\text{Limits Do Not Exist}\\ \underset{x\to \pm \mathrm{\infty }}{lim}\tan (x)\underset{x\to \pm \mathrm{\infty }}{lim}\cot (x)\text{Limits Do Not Exist}\\ \underset{x\to \pm \mathrm{\infty }}{lim}\sec (x)\underset{x\to \pm \mathrm{\infty }}{lim}\csc (x)\text{Limits Do Not Exist}\end{array}$$

  
  ::==================================================================x=============================================================================================x===============================================================不===================不==============

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  Let's find find $\begin{array}{r}\underset{x\to 0}{lim}f(x)\end{array}$  for $\begin{array}{r}f(x)=x^2\cos (10\pi x)\end{array}$
  ::让我们找到 f( x) =x2cos *( 10x) 的limx% 0f( x) 。

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  The graph of the function is shown below.
  ::函数图如下。

  

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  Since we know that the limit of  $\begin{array}{r}{x}^2\end{array}$ and  $\begin{array}{r}\cos (x)\end{array}$ exist, we can find the limit of this function by applying the Product Rule, or direct substitution:
  ::由于我们知道x2和cos(x)的限值存在,我们可以通过适用产品规则或直接替代来找到这一功能的限值:

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  $$\begin{array}{rl}\underset{x\to 0}{lim}{x}^2\cos (10\pi x)& =(\underset{x\to 0}{lim}{x}^2)\cdot (\underset{x\to 0}{lim}\cos (10\pi x))\\ & =0\cdot 1\\ & =0\end{array}$$

  
  :10xx) =( limx=0x2) =( limx=0x2) =( limx=0x) =( 10xx) =( 0xx) =( 0xxx) =( 0xx) =( 0xx) =( 0xxx) =( 0xx)

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  Hence, $\begin{array}{r}\underset{x\to 0}{lim}{x}^2\cos (10\pi x)=0\end{array}$ .
  ::因此, limx0x2cos( 10xx) =0。

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  Also, from the graph of the function, we note that the function is bounded by the graphs of  $\begin{array}{r}{x}^2\end{array}$ and $\begin{array}{r}-x^2\end{array}$ , which both are 0 at $\begin{array}{r}x=0\end{array}$ . It makes sense that the original function’s limit should be 0.
  ::另外,从函数的图形中,我们注意到,函数受x2和-x2的图形所约束,它们都在x=0时为0。 原始函数的极限应该是0是有道理的。

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  The Squeeze Theorem
  ::震动定理

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  This feature, that a function’s limit can result from the function being bounded or squeezed by two other functions, is the basis for the Squeeze Theorem . The Squeeze Theorem (also known as the Sandwich Theorem) states:
  ::这个特性,函数的极限可能来自被另外两个函数捆绑或挤压的函数,这是“挤压定理”的基础。

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  If $\begin{array}{r}f(x)\le g(x)\le h(x)\end{array}$  for every  $\begin{array}{r}x\end{array}$ in an open interval containing $\begin{array}{r}a\end{array}$ , and $\begin{array}{r}\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}h(x)=L\end{array}$ , then $\begin{array}{r}\underset{x\to a}{lim}g(x)=L\end{array}$ .
  ::如果 f(x)\g(x)\h(x)x 在开放间隔内每个 x 都包含 a, 和 limx=limxah(x) =L, 那么 limx)\g(x) =L 。

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  In other words, if we can find bounds for a function that have the same limit, then the limit of the function that they bound must have the same limit.  Note that $\begin{array}{r}a\end{array}$ and $\begin{array}{r}L\end{array}$ may be any constant or even  $\begin{array}{r}\mathrm{\infty }\end{array}$ or $\begin{array}{r}-\mathrm{\infty }\end{array}$ .
  ::换句话说,如果我们能找到一个具有相同限制的函数的界限,那么它们约束的函数的界限必须具有相同的界限。请注意,a和L可能是任何恒定的,甚至或。

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  One of the important trigonometric limits that can be proved, in part, using the Squeeze Theorem is:
  ::可以部分地证明,使用 " 压强定理 " 的重要三角界限之一是:

  $$\begin{array}{r}\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)=1\end{array}$$

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  Another important trigonometric limit is   $\begin{array}{r}\underset{x\to 0}{lim}\left(\frac{1-\cos x}{x}\right)\end{array}$ . 
  ::另一个重要的三角限值是 limx0(1-cosxxxxx)。

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  Direct substitution cannot be used to evaluate the limit because it yields the indeterminate form $\begin{array}{r}\frac{0}{0}\end{array}$ . Instead, transform the problem to a different form and solve.
  ::不能使用直接替代来评估限额,因为它产生不确定的00表。 相反,它会将问题转换成另一种形式并解决问题。

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  $$\begin{array}{rl}\underset{x\to 0}{lim}\left(\frac{1-\cos x}{x}\right)& =\underset{x\to 0}{lim}\left(\frac{1-\cos x}{x}\cdot \frac{1+\cos x}{1+\cos x}\right)\\ & =\underset{x\to 0}{lim}\left(\frac{1-{\cos }^{2}x}{x(1+\cos x)}\right)\\ & =\underset{x\to 0}{lim}\left(\frac{{\sin }^{2}x}{x(1+\cos x)}\right)\\ & =\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\cdot \frac{\sin x}{1+\cos x}\right)\\ & =\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)\cdot \underset{x\to 0}{lim}\left(\frac{\sin x}{1+\cos x}\right)\\ & =(1)\left(\frac{0}{2}\right)\\ \underset{x\to 0}{lim}\left(\frac{1\cos x}{x}\right)& =0\end{array}$$

  
  ::===0 =0 =1 =1 =1 =1 =1 =2xxxx(1+2xxxx) = limx0xxxxx(1+2xxxxx(1+xxxxxx) = limx0x0(sinxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1+xxxxxxxxxxxxxxxxxxxxxxxxxxx=1 02limxxxxxxxxxx(1+xxxxxxxxxxxxxxx) = limxx0xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked if  the periodic nature of trigonometric functions and the limited or infinity range of individual trigonometric functions make evaluating limits involving trigonometric  functions difficult.
  ::早些时候,有人问您,三角函数的周期性以及个别三角函数的有限或无限范围是否使得评估涉及三角函数的限制有困难。

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  As you can imagine and have seen in this concept, some limits involving trigonometric functions can be easily evaluated by direct substitution, and some evolve a lot of work to change form from an indeterminate or undefined form. Determining the end behavior of an expression involving a trigonometric function can also be difficult, and require application of principles like the Squeeze Theorem to obtain a result. No easy answer!
  ::正如你们可以想象和从这个概念中已经看到的那样,涉及三角函数的某些限制可以通过直接替代来轻易地评估,而有些则会从不确定或未定义的形式改变许多工作。 确定涉及三角函数的表达方式的最终行为也可能很困难,并且需要应用“震动理论”等原则才能获得结果。 没有简单答案!

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  Example 2
  ::例2

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  Find $\begin{array}{r}\underset{x\to 0}{lim}{x}^2\cos \left(\frac{10\pi }{x}\right)\end{array}$ .
  ::查找 limx=% 0x2cos {} (10xx) 。

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  Direct substitution cannot be used to evaluate the limit because $\begin{array}{r}\frac{10\pi }{x}\end{array}$  is undefined when $\begin{array}{r}x=0\end{array}$ .
  ::不能使用直接替代来评价限制, 因为 10x 在 x=0 时没有定义 。

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  However, the Squeeze Theorem can be used as follows:
  ::然而,可使用以下的 " 压强定理 " :

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  1. We know that cosine stays between -1 and 1, so $\begin{array}{r}-1\le \cos \left(\frac{10\pi }{x}\right)\le 1\end{array}$  for any  $\begin{array}{r}x\end{array}$ in the domain of the function (i.e., any $\begin{array}{r}x\ne 0\end{array}$ ).
     ::我们知道余弦在 -1 和 1 之间, 所以- 1 cos( 10xx) 1 用于函数范围内的任何 x (即任何 xx) 。
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  2. Since $\begin{array}{r}{x}^2\end{array}$  is always non-negative, we can multiply the above inequality by $\begin{array}{r}{x}^2\end{array}$ : $\begin{array}{r}-x^2\le \cos \left(\frac{10\pi }{x}\right)\le x^2\end{array}$ .
     ::由于 x2 总是非负性的, 我们可以将上述不平等乘以 x2 : - x2cos( 10x) x2 。
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  3. The original function is bounded by $\begin{array}{r}{x}^2\end{array}$  and  $\begin{array}{r}-x^2\end{array}$ and $\begin{array}{r}\underset{x\to 0}{lim}-x^2=\underset{x\to 0}{lim}{x}^2=0\end{array}$ .
     ::原函数受 x2 和 ~x2 和 limx= 0 - x2 = limx= 0x2 = 0 的约束。
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  4. Therefore, by the Squeeze Theorem: $\begin{array}{r}\underset{x\to 0}{lim}{x}^2\cos \left(\frac{10\pi }{x}\right)=0\end{array}$ .
     ::因此,以震动定理: limx0x2cos(10xx)=0。

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  Review
  ::回顾

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  1. Find $\begin{array}{r}\underset{x\to \frac{\pi }{6}}{lim}\sin (x)\end{array}$
     ::查找 limx6sin(x)
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  2. Find $\begin{array}{r}\underset{x\to \frac{\pi }{4}}{lim}\cot (x)\end{array}$
     ::查找 limx4cot(x)
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  3. Find $\begin{array}{r}\underset{x\to \frac{\pi }{3}}{lim}{\sec }^2(x)\end{array}$
     ::查找 limx3sec2\(x)
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  4. Find $\begin{array}{r}\underset{x\to \frac{\pi }{3}}{lim}[\sin (x)+\cos (x)]\end{array}$
     ::查找 limx3 [sin(x)+cos(x)]
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  5. Find $\begin{array}{r}\underset{x\to \frac{\pi }{2}}{lim}[\sec (x)+\tan (x)]\end{array}$
     ::查找 limx @%% 2 [sec}( x) +tan} (x)]
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  6. Find $\begin{array}{r}\underset{x\to \pi }{lim}\frac{\sin x}{\cos x+1}\end{array}$
     ::查找 limx =%sin =xcos=x+1 查找 limx =xcos=x+1
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  7. Find $\begin{array}{r}\underset{x\to 0}{lim}\frac{\sin (x)}{3x}\end{array}$
     ::查找 limx0sin(x)3x
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  8. Find $\begin{array}{r}\underset{x\to 0}{lim}\frac{2\cos (x)-2}{x}\end{array}$
     ::查找 limx02cos(x)- 2x
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  9. Find $\begin{array}{r}\underset{x\to 0}{lim}\frac{\sin (3x)}{x}\end{array}$
     ::查找 limx0sin( 3x) x
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  10. Find $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim}x\sin \left(\frac{1}{x}\right)\end{array}$
      ::查找 limxxsin( 1x)
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  11. Find $\begin{array}{r}\underset{x\to 1}{lim}\frac{\cos (x^2-1)-1}{x^2-1}\end{array}$
      ::查找 limx% 1cos {( x2 - 1) - 1x2 - 1
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  12. Find $\begin{array}{r}\underset{x\to 0}{lim}\frac{\cos (x)}{x}\end{array}$
      ::查找 limx%%0cos @(x) x
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  13. Find $\begin{array}{r}\underset{x\to 0}{lim}(2x\cot 2x)\end{array}$
      ::查找 limx%% 0( 2xcot%%% 2x)
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  14. Find $\begin{array}{r}\underset{x\to -\frac{3}{2}}{lim}\left[\frac{\sin (2x+3)}{2x^2+x-3}\right]\end{array}$
      ::查找 limx% 32 [sin( 2x+3) 2x2+x-3]
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  15. Find $\begin{array}{r}\underset{x\to 0}{lim}\left[\frac{\tan x}{x}\right]\end{array}$
      ::查找 limx% 0 [tanxxxx]

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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 。