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求极限

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APPSYZY Posted at 2020-6-2 14:30:49 |Read mode
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 Author| APPSYZY Posted at 2020-6-2 14:32:32
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这应该是一个错误的解答

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 Author| APPSYZY Posted at 2020-6-2 14:33:13
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这应该是也是错误的

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 Author| APPSYZY Posted at 2020-6-2 14:34:17
错误的原因是,题目并没有告诉f'(x)在x=1处连续,所以不能乱用洛必达法则,具体怎么解还请高人指点。

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kuing Posted at 2020-6-2 14:45:48
\begin{align*}
\text{原式}&=\lim_{x\to0}\frac{f(\sin^2x+\cos x)-f(1)}{\sin^2x+\cos x-1}\cdot\frac{(\sin^2x+\cos x-1)\tan4x}{(e^{x^2}-1)\sin x}\\
&=f'(1)\cdot\lim_{x\to0}\frac{\cos x\tan(x/2)\tan4x}{e^{x^2}-1}\\
&=f'(1)\cdot\lim_{x\to0}\frac{2x^2}{e^{x^2}-1}\\
&=2f'(1).
\end{align*}

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