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来自人教群的骗过软件的极限

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kuing Posted at 2014-10-21 00:31:20 |Read mode
爱好者-jqs271828(6768*****)  23:58:27
limit (e^x+sinx)/(e^x-sinx),x->-∞

\[\lim_{x\to-\infty}\frac{e^x+\sin x}{e^x-\sin x}.\]

记 $f(x)=(e^x+\sin x)/(e^x-\sin x)$,显然 $f(-n\pi)=1$;

又设 $e^x+\sin x=0$ 的所有零点由大到小排列为数列 $\{x_n\}$,易见 $n\to+\infty$ 时 $x_n\to-\infty$,显然恒有 $\sin x_n\ne0$,故 $e^{x_n}-\sin x_n\ne0$,即恒有 $f(x_n)=0$。

综上知极限不存在。

但是软件就被骗过了:
QQ截图20141021003057.gif

不知你们用的其他软件如何呢?

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caijinzhi Posted at 2014-10-23 00:08:01
能否借此探索软件求极限的方法?

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战巡 Posted at 2014-10-23 07:03:39
回复 2# caijinzhi


你直接带入x=-∞就会得到-1了

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hbghlyj Posted at 2023-2-19 00:05:34
SageMath输出0
Screenshot 2023-02-18 at 16-05-09 Sage Cell Server.png

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hbghlyj Posted at 2023-2-19 00:56:22
WolframAlpha是正确的

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hbghlyj Posted at 2023-2-19 00:59:06
Matlab
  1. syms x;
  2. y = (exp(x) + sin(x)) / (exp(x) - sin(x));
  3. limit(y, x, -inf)
Copy the Code

ans = $-1$

Octave
Screenshot 2023-02-18 at 17-00-33 Octave Online · Cloud IDE compatible with MATLAB.png

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hbghlyj Posted at 2023-5-8 05:07:00

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hbghlyj Posted at 2023-5-15 23:20:47
知乎—为啥wolfram alpha不但判断错了tan(n)/n的敛散性,还给出了一个匪夷所思的证明过程?
WolframAlpha
Untitled.png

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2025-4-20 22:28 GMT+8

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