洛必达法则是这样证明的，为什么f(a),g(a)可以假设等于0？数学分析 高数
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由2楼，所以说树上的证明不完全，但是是正确的，或者说应该在某个范围内是正确的，教材内容是不会超过这个范围的，我觉得楼主是陷入这样一个思维误区：特殊性是证明不出一般性的，但是逻辑告诉我们，这是可行的，所以事实上楼主应该是逻辑混乱了，我复制粘贴了一下2楼的，也就是一般性证明？。General proof
The following proof is due to (Taylor 1952), where a unified proof for the 0/0 and ±∞/±∞ indeterminate forms is given. Taylor notes that different proofs may be found in (Lettenmeyer 1936) and (Wazewski 1949).
Let f and g be functions satisfying the hypotheses in the General form section. Let \mathcal{I} be the open interval in the hypothesis with endpoint c. Considering that g'(x)\neq 0 on this interval and g is continuous, \mathcal{I} can be chosen smaller so that g is nonzero on \mathcal{I}.[6]
For each x in the interval, define m(x)=\inf\frac{f'(\xi)}{g'(\xi)} and M(x)=\sup\frac{f'(\xi)}{g'(\xi)} as \xi ranges over all values between x and c. (The symbols inf and sup denote the infimum and supremum.)
From the differentiability of f and g on \mathcal{I}, Cauchy's mean value theorem ensures that for any two distinct points x and y in \mathcal{I} there exists a \xi between x and y such that \frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(\xi)}{g'(\xi)}. Consequently m(x)\leq \frac{f(x)-f(y)}{g(x)-g(y)} \leq M(x) for all choices of distinct x and y in the interval. The value g(x)-g(y) is always nonzero for distinct x and y in the interval, for if it was not, the mean value theorem would imply the existence of a p between x and y such that g' (p)=0.
The definition of m(x) and M(x) will result in an extended real number, and so it is possible for them to take on the values ±∞. In the following two cases, m(x) and M(x) will establish bounds on the ratio f/g.
The following proof is due to (Taylor 1952), where a unified proof for the 0/0 and ±∞/±∞ indeterminate forms is given. Taylor notes that different proofs may be found in (Lettenmeyer 1936) and (Wazewski 1949).
Let f and g be functions satisfying the hypotheses in the General form section. Let \mathcal{I} be the open interval in the hypothesis with endpoint c. Considering that g'(x)\neq 0 on this interval and g is continuous, \mathcal{I} can be chosen smaller so that g is nonzero on \mathcal{I}.[6]
For each x in the interval, define m(x)=\inf\frac{f'(\xi)}{g'(\xi)} and M(x)=\sup\frac{f'(\xi)}{g'(\xi)} as \xi ranges over all values between x and c. (The symbols inf and sup denote the infimum and supremum.)
From the differentiability of f and g on \mathcal{I}, Cauchy's mean value theorem ensures that for any two distinct points x and y in \mathcal{I} there exists a \xi between x and y such that \frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(\xi)}{g'(\xi)}. Consequently m(x)\leq \frac{f(x)-f(y)}{g(x)-g(y)} \leq M(x) for all choices of distinct x and y in the interval. The value g(x)-g(y) is always nonzero for distinct x and y in the interval, for if it was not, the mean value theorem would imply the existence of a p between x and y such that g' (p)=0.
The definition of m(x) and M(x) will result in an extended real number, and so it is possible for them to take on the values ±∞. In the following two cases, m(x) and M(x) will establish bounds on the ratio f/g.
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这样假设了就可以应用柯西定理了，好像极限里有时补个可去间断点就可以用拉格朗日定理了，明白？
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但是f(a),g(a)不一定等于0啊
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你把他设成几都无所谓，因为不影响，那不如干脆都设成零，这样就满足了柯西定理，就可以简化运算了。你考研？
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【关于洛必达法则证明的几点补充 - 豆丁网】
https://www.docin.com/touch/detail.do?id=1033849799
https://www.docin.com/touch/detail.do?id=1033849799
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看一看数学分析
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