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怎么求这个极限$\lim_{x \to 1}\frac {(x-2)e^x}{(x-1)^2}$

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isee posted 2017-4-9 10:51 |Read mode
这样有理论依据么?
$$\lim_{x \to 1}\frac {(x-2)e^x}{(x-1)^2}=\lim_{x \to 1}(x-2)e^x\lim_{x \to 1}\frac 1{(x-1)^2}=-e\cdot (+\infty)=-\infty?$$

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abababa posted 2017-4-9 11:11
回复 1# isee

可以,就是积的极限等于极限之积,不是$0\cdot\infty$的未定式,而且包含广义实数的情况下极限是$\infty$也是存在,不像$\lim_{x\to\infty}\sin x$这样的是不存在。

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original poster isee posted 2017-4-9 11:28
回复  isee

可以,就是积的极限等于极限之积,不是$0\cdot\infty$的未定式,而且包含广义实数的情况下极 ...
abababa 发表于 2017-4-9 11:11
这种极限困惑了我很久,说到我不明白地方上了,谢谢。。。

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其妙 posted 2017-4-9 23:28
回复 1# isee
对的

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