课程： 4.7 评估不确定限度:医院规则

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区
选择节评估不确定限度:医院规则

折叠展开
评估不确定限度:医院规则
Often in the evaluation of the limit of a function, the use of the direct substitution method leads to an indeterminate form such as $\begin{aligned} \frac{0}{0} \end{aligned}$ or $\begin{aligned} \frac{\infty}{\infty} \end{aligned}$ , and another method must be tried. Do you remember the factor canceling method used on some rational expressions (polynomial numerator and denominator), and the conjugate multiplication method used on some radical expressions? The method introduced in this concept transforms the original indeterminate form by taking the derivative of the numerator and denominator of the indeterminate form, and then applying the limit again. Why would such technique would provide a better chance to find the limit?
::通常在评估函数的极限时, 直接替代方法的使用导致一种不确定的形式, 如 00 或或 , 以及另一种方法必须尝试。您还记得某些理性表达式( 极分子和分母) 所使用的取消因子的方法, 以及某些激进表达式所使用的同源乘法吗 ? 这一概念中引入的方法通过取取定数和分母的衍生物来改变原不确定的形式, 然后再次应用该限制。为什么这种技术能提供更好的机会来找到极限呢 ?

L'Hôpital's Rule
::“医院规则”

The primary indeterminate forms that are applicable for using the method we will be discussing here are:
::适用于使用我们在这里讨论的方法的主要不确定形式是:

$\begin{aligned} \frac{0}{0}, \frac{\infty}{\infty}, \frac{- \infty}{\infty}, \frac{\infty}{- \infty}, \frac{- \infty}{- \infty} \end{aligned}$

As an example, $\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} \end{aligned}$ cannot be evaluated by direct substitution because the result is the indeterminate form $\begin{aligned} \frac{0}{0} \end{aligned}$ .
::例如,不能通过直接替代来评价limx0sinxxxxx,因为结果为不确定的表格00。

There is a method that can be used, in many cases, to evaluate limits when direct substitution or other methods result in an indeterminate form.
::在许多情况下,当直接替代或其他方法造成不确定的形式时,可以使用一种方法来评价限制。

This method is called L’Hôpital’s Rule:
::这个方法叫做医院规则:

Let functions $\begin{aligned} f \end{aligned}$ and $\begin{aligned} g \end{aligned}$ be differentiable at every number other than $\begin{aligned} c \end{aligned}$ in some interval, with $\begin{aligned} g^{'} (x) \neq 0 \end{aligned}$ if $\begin{aligned} x \neq c . \end{aligned}$
::Let f 和 g 函数在除 c 以外的每个数字中都有差异, 如果 x @ c, 则使用 g_ (x) @% 0 。

If $\begin{aligned} lim_{x \to c} f (x) = 0 \end{aligned}$ and $\begin{aligned} lim_{x \to c} g (x) = 0 \end{aligned}$ , or if $\begin{aligned} lim_{x \to c} f (x) = \pm \infty \end{aligned}$ and $\begin{aligned} lim_{x \to c} g (x) = \pm \infty \end{aligned}$ , then:
::如果(x)=0和(x)=0,或者(x)=0,或者(x)=% 和(x)=% 和(x)=%,那么:

$\begin{aligned} lim_{x \to c} \frac{f (x)}{g (x)} = lim_{x \to c} \frac{f^{'} (x)}{g^{'} (x)} \end{aligned}$

:x)g(x)=limx=c#(x)g(x)

as long as $\begin{aligned} lim_{x \to c} \frac{f^{'} (x)}{g^{'} (x)} \end{aligned}$ exists or is infinite.
::只要limx{cf}(x)g}(x)存在或无限。

If $\begin{aligned} f \end{aligned}$ and $\begin{aligned} g \end{aligned}$ are differentiable at every number $\begin{aligned} x \end{aligned}$ greater than some number $\begin{aligned} a \end{aligned}$ , with $\begin{aligned} g^{'} (x) \neq 0 \end{aligned}$ then:
::如果 f 和 g 在每一数字x 大于某个数字a 时是可区分的, 则 g( x) @% 0 然后 :

$\begin{aligned} lim_{x \to \pm \infty} \frac{f (x)}{g (x)} = lim_{x \to \pm \infty} \frac{f^{'} (x)}{g^{'} (x)} \end{aligned}$

::limxf( x) g( x) =limx\\\ f}( x) g_( x) (x)

as long as $\begin{aligned} lim_{x \to \pm \infty} \frac{f^{'} (x)}{g^{'} (x)} \end{aligned}$ exists or is infinite.
::只要limxf(x)g(x)存在或无限。

Note that L’Hôpital’s rule is also valid for .
::请注意,L ' hital的规则也适用于...。

Let’s look again at finding $\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} \end{aligned}$ using L’Hôpital’s Rule.
::让我们再看看使用医院规则来寻找limx0sinxxxxxx。

Find $\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} \end{aligned}$ .
::找找 lix_0sinxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

For this problem $\begin{aligned} f (x) = \sin x \end{aligned}$ and $\begin{aligned} g (x) = x \end{aligned}$ .
::对于这个问题 f( x) =sinx 和 g( x) =x 。

Since $\begin{aligned} lim_{x \to 0} \sin x = lim_{x \to 0} x = 0 \end{aligned}$ , L’Hôpital’s Rule applies and we have
::自Limx0sinx=limx0x=0,医院的规则适用以来,我们已经有了

$\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} & = lim_{x \to 0} \frac{\frac{d}{d x} [\sin x]}{\frac{d}{d x} [x]} \\ = lim_{x \to 0} \frac{\cos x}{1} \\ = \frac{1}{1} \\ = 1 \end{aligned}$

::立方公尺xxxx= 立方公尺xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1=11=1xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1=1

Therefore, $\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} = 1 \end{aligned}$ .
::因此,limx=0sin=xxxx=1。

There are some limits where L’Hôpital’s Rule must be applied more than once before a determinate form results.
::有些限制是,医院的规则在确定结果之前必须不止一次适用一次。

Let's take $\begin{aligned} lim_{x \to + \infty} \frac{x^{2}}{e^{x}} \end{aligned}$ .
::来来来来来来来来来来来来来去去去去去去去

Direct substitution results in an indeterminate form of $\begin{aligned} \frac{\infty}{\infty} \end{aligned}$ .
::直接替代导致一种不确定的________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Since $\begin{aligned} lim_{x \to + \infty} x^{2} = lim_{x \to + \infty} e^{x} = + \infty \end{aligned}$ , L’Hôpital’s Rule applies and we have:
::自Limxx2=limxxex, 医院的规则适用,

$\begin{aligned} lim_{x \to + \infty} \frac{x^{2}}{e^{x}} & = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x}]} \\ lim_{x \to + \infty} \frac{x^{2}}{e^{x}} & = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x}]} \\ = lim_{x \to \infty} \frac{2 x}{e^{x}} \dots Notice that applying L'H \hat{o} pital's Rule has resulted \\ in the same indeterminate form \frac{\infty}{\infty} . \\ Try\ L'H \hat{o} pital's\ Rule\ again! \\ = lim_{x \to \infty} \frac{2}{e^{x}} \ldots This can be evaluated. \\ = 0 \end{aligned}$

::注意应用L'医院规则已产生相同的不确定形式。Try\ L'医院规则\\\\\!=limx\\\\2\ex\\ld\ldots\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Therefore, $\begin{aligned} lim_{x \to + \infty} \frac{x^{2}}{e^{x}} = 0 \end{aligned}$ .
::因此, limxx2ex=0。

L’Hôpital’s Rule can be used repeatedly on functions like this. It is often useful because polynomial functions can be reduced to a constant.
::医院规则可以反复用于类似功能。它常常有用,因为多面性功能可以减为常数。

The indeterminate forms listed above are the ones that apply directly to the use of L’Hôpital’s Rule, but those are not the only indeterminate forms.
::上面列出的不确定形式是直接适用于使用医院规则的形式,但这些形式并非唯一的不确定形式。

Other indeterminate forms such as the following can occur when evaluating limits:
::在评估限额时,还可能出现其他不确定的形式,例如以下形式:

$\begin{aligned} 0 \cdot \pm \infty \infty - \infty 0^{0} \infty^{0} 1^{\infty} \end{aligned}$

These other forms are not applicable using L’Hôpital’s Rule, but can sometimes be transformed to one of the four indeterminate forms that is.
::这些其他表格不适用《医院规则》,但有时可以转换成四种不确定的形式之一。

Take $\begin{aligned} lim_{x \to 0^{+}} \sqrt{x} \cdot \ln x \end{aligned}$ for example.
::例如, limx=0+x=xx=%x 。

Evaluating the limit using direct substitution yields the indeterminate for $\begin{aligned} 0 \cdot - \infty \end{aligned}$ , which is not one of the forms applicable to L’Hôpital’s Rule. The remedy is to convert the limit product to a limit quotient:
::评估使用直接替代的限值,得出0的不确定值,这不是适用于医院规则的形式之一。补救措施是将限值产品转换为限值:

$\begin{aligned} lim_{x \to 0^{+}} \sqrt{x} \cdot \ln x & = lim_{x \to 0^{+}} \frac{\ln x}{\frac{1}{\sqrt{x}}} \ldots Current form must change for use with L'H \hat{o} pital's Rule : \\ Convert the product to a quotient. \\ = lim_{x \to 0^{+}} \frac{\frac{d}{d x} \ln x}{\frac{d}{d x} [\frac{1}{\sqrt{x}}]} \ldots Apply L'H \hat{o} pital's Rule \\ = lim_{x \to 0^{+}} \frac{\frac{1}{x}}{- \frac{1}{2} x^{\frac{- 3}{2}}} \\ = lim_{x \to 0^{+}} [- 2 \sqrt{x}] \\ = 0 \end{aligned}$

::=limx=0+ddxln=xddx[1x]\ldots 应用 L'hods's rules=limx_0+1x-12x-32=limx=0+[ -2x]=0

Therefore, $\begin{aligned} lim_{x \to 0^{+}} \sqrt{x} \cdot \ln x = 0 \end{aligned}$ .
::因此, limx=0+xx=0=0=0=xx=0=xxx=0=xxx=0=xxxxx=0=xxxxxxx=0=xxxxxxxxxxx=0=xxxxxxxxxxxxxxx=0=xxxxxxxxxx=0=xxxxxxxxxx=0=xxxxxxxxxxxx=0=xxxxxxxxxxxxx=0=xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx=0=0=0=xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Examples
::实例

Example 1
::例1

Earlier, you were asked why taking the derivative of numerator and denominator of the limit form provides a better chance to find the limit.
::早些时候,有人问过,为什么使用限量表的分子和分母的衍生物, 提供了更好的机会找到限值。

Apart from a proof of L’Hôpital’s Rule, there are probably many intuitive reasons for why it works. Perhaps you said that the rate of change of the numerator versus the denominator is important in determining the ratio. If the denominator is getting to infinity (0) faster than the denominator, that information matters, and the derivative gives it.
::除了证明医院规则外,它之所以起作用可能有许多直觉原因。也许你说过,分子相对于分母的变动率对于确定比率很重要。如果分母比分母(0)的无限速度快,信息很重要,衍生物也给信息带来好处。

Example 2
::例2

Evaluate $\begin{aligned} lim_{x \to 0} \frac{e^{x} - 1}{x} . \end{aligned}$
::评估 limx0ex -1x。

Direct substitution results in an indeterminate form.
::直接替代导致一种不确定的形式。

Since $\begin{aligned} lim_{x \to 0} (e^{x} - 1) = lim_{x \to 0} x = 0 \end{aligned}$ , L’Hôpital’s Rule applies and we have
::自Limx%%0(ex-1)=limx=0=0, 医院规则适用,我们已执行

$\begin{aligned} lim_{x \to 0} \frac{e^{x} - 1}{x} & = lim_{x \to 0} \frac{\frac{d}{d x} [e^{x} - 1]}{\frac{d}{d x} [x]} \\ = lim_{x \to 0} \frac{e^{x}}{1} \\ = \frac{1}{1} \\ = 1 \end{aligned}$

::limx_0ex- 1x=limx%0dx[ex-1]ddx[x]=limx_0ex1=11=1]

Therefore, $\begin{aligned} lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 \end{aligned}$ .
::因此, limx=0ex- 1x=1。

Example 3
::例3

Evaluate $\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} \end{aligned}$ .
::评估 limx01-cosxxxxxxxxxxxxxxxxxxxxxxxxxlxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Direction substitution results in the indeterminate form $\begin{aligned} \frac{0}{0} \end{aligned}$ .
::方向替代结果为不确定型号为00。

Since $\begin{aligned} lim_{x \to 0} (1 - \cos x) = lim_{x \to 0} x^{2} = 0 \end{aligned}$ , L’Hôpital’s Rule applies and we have
::自Limx%0( 1 - cosx) = limx=0x2=0, 医院的规则适用, 我们已经使用

$\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} & = lim_{x \to 0} \frac{\frac{d}{d x} [1 - \cos x]}{\frac{d}{d x} [x^{2}]} \\ = lim_{x \to 0} \frac{\sin x}{2 x} \ldots Notice that applaying L'H \hat{o} pital's Rule has resulted \\ in the same indeterminate form \frac{\infty}{\infty} . \\ Try\ L'H \hat{o} pital's\ Rule\ again! \\ = lim_{x \to 0} \frac{\cos x}{2} \ldots This limit can be evaluated . \\ = \frac{1}{2} . \end{aligned}$

::==limx=0cos=x2=limx=0sin=x2xxxxdx[x2]ddx[x2]=limxx=0sin}x2x\ldots] 提醒注意,对医院规则的适应产生了相同的不确定形式。尝试\ L'医院规则\ 规则\ 重现!=limx=0cos=x2\ldots 此限制可以被评估 =12。

Therefore, $\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2} . \end{aligned}$
::因此, limx01 -cosxx2=12。

Review
::回顾

For all problems, use L’Hôpital’s Rule to compute the limits, if they exist.
::对于所有问题,如果存在限制,请使用医院规则来计算限制。

$\begin{aligned} lim_{x \to 3} \frac{x^{2} - 9}{x - 3} \end{aligned}$
::limx=3x2-9x-3

$\begin{aligned} lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} \end{aligned}$
::limx=01+x-1-xxxxxxxxxxxxx

$\begin{aligned} lim_{x \to + \infty} \frac{\ln (x)}{\sqrt{x}} \end{aligned}$
::立方公尺 [x]x

$\begin{aligned} lim_{x \to + \infty} x^{2} e^{- 2 x} \end{aligned}$
::limxx2e-2x

$\begin{aligned} lim_{x \to 0} (1 - x)^{\frac{1}{x}} \end{aligned}$
::limx0( 1- x) 1x

$\begin{aligned} lim_{x \to 0} \frac{e^{x} - 1 - x}{x^{2}} \end{aligned}$
::立方厘米x0ex-1-2x2

$\begin{aligned} lim_{x \to - \infty} \frac{e^{x} - 1 - x}{x^{2}} \end{aligned}$
::立方公尺xex - 1 - xx2

$\begin{aligned} lim_{x \to \frac{π}{2}} \frac{\tan x}{1 + \tan x} \end{aligned}$
::立方公尺xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1+tan1+xxxxxx

$\begin{aligned} lim_{x \to 0} \frac{\tan x}{x} \end{aligned}$
::{\fn方正黑体简体\fs18\b1\bord1\shad1\3cH2F2F2F

$\begin{aligned} lim_{x \to 0} \frac{x^{2}}{\sin^{2} x} \end{aligned}$
::limx=0x2sin2x limx=0x2sin2x

$\begin{aligned} lim_{x \to 0} \frac{e^{x^{2}} - 1}{\sin x^{2}} \end{aligned}$
::立方公尺x% 0ex2 - 1sin%x2

$\begin{aligned} lim_{x \to \infty} \frac{\sqrt{x^{2} + 1}}{x} \end{aligned}$
::limxx2+1x

$\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} \end{aligned}$
::立方公尺

$\begin{aligned} lim_{x \to \infty} x^{3} e^{- x} \end{aligned}$
::limxx3e-x

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。

章节大纲

常规

评估不确定限度:医院规则

L'Hôpital's Rule ::“医院规则”

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

L'Hôpital's Rule
::“医院规则”

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)