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在一个微积分读本上看到的,求:
\(\lim\limits_{h\rightarrow 0}\frac{\ln(1-7h^2)}{5h^2}\)
书上的答案是,分母上换成\(-7h^2\),即:
\(\lim\limits_{h\rightarrow 0}\frac{\ln(1-7h^2)}{5h^2}=\lim\limits_{h\rightarrow 0}\frac{\ln(1-7h^2)}{-7h^2}\times\frac{-7h^2}{5h^2}=\frac{-7}{5}\)
但是,原式子要转成:\(\lim\limits_{-7h^2\rightarrow 0}\frac{\ln(1-7h^2)}{-7h^2}\times\frac{-7h^2}{5h^2}\)才对,\(-7h^2恒小于等于0\),上式只有左极限,没有右极限,极限值应该不存在的,怎么会求出值呢?
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