WEBVTT

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我们想求出随着x接近1时，表达式x/x-1

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乘上1/ln x的极限值

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所以让我们看下当我们仅是输入1时

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会发生什么

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好，接着我们在此处得到1，1-1

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所以我们会得到1/0减去

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1除以，1的自然对数是多少呢

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e的几次方等于1呢

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任何数的零次幂都为1，所以e的零次幂也为1

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所以1的自然对数值为

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0

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所以我们得到了奇怪且无解的1/0-1/0

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这是一种奇怪的无解形式

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但这并不是我们在l'Hopital's rule 中看到的无解形式

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我们不会求出0/0，也不会求出∞/∞

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所以你也许会说，好吧，这不是一个l'Hopital 法则的问题

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我们须以另一种方式将此题解出

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不要放弃呀

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也许我们可以以某种代数的方式改写这个式子

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以使其变化为l'Hopital 的不确定形式

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接着我们就可将之直接运用了

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为了将之解决，让我们看看如果将这两式相加

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又会如何呢

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所以如果我将这两式相加

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分母将为（x-1）*ln x

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我仅是将这两式相乘

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接着分子将为

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好的，如果我将这整个式子同时乘上ln x

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所以分子将为x*ln x

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这个式子我将其整体乘以（x-1）

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那么即为-（x-1）

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你可以将之拆分（验证），并发现其与原式一致

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那么在这边，x/x-1,由于ln x消掉了

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让我们将其搁一边

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那么这边就是-1/ln x由于（x-1）被消掉了

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希望你理解了关于我处理这两表达式的用意

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所以借此入手，我们看下取x为1时

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此式会如何变化呢

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因为这些式子是相同的

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所以我们得到什么了吗

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我们得到了1*ln 1

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由于ln 1的值为0所以我们就得到了0

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减去0，所以原式值为0

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所以我们有了0作为分子

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并且在分母中，我们求出了1-1，值为零乘上

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ln 1， 其值也为0，故分母值也为0

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我们得到了应用l'Hopital法则所需的不定形式

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