强形式洛必达法则

(L’Hospital law) Suppose \(f\colon (a,b)\rightarrow \mathbb R\) and \(g\colon(a,b)\rightarrow \mathbb R\) are differiential in \((a,b)\) (\(-\infty\le a<b\le +\infty\)). \(g'(x)\ne 0\) in \((a,b)\) and

\[\dfrac{f(x)}{g(x)}\rightarrow A,\ \ x\rightarrow a^{+}\ (-\infty\le A\le +\infty). \]

When either (1) or (2) given by

(1) \(f(x)\rightarrow 0\) and \(g(x)\rightarrow 0\) as \(x\rightarrow a^{+}\).

(2) \(g(x)\rightarrow \infty\) as \(x\rightarrow a^{+}\).

happens, we have

\[\dfrac{f(x)}{g(x)}\rightarrow A,\ \ x\rightarrow a^{+}. \]

We can rewrite this result by replacing \(x\rightarrow a^{+}\) by \(x\rightarrow b^{-}\).

Proof: Since \(g'(x)\ne 0\), \(g(x)\) is strictly monotonous in \((a,b)\). Thus when either (1) or (2) happens, there exists \(\delta>0\) such that \(g(x)\ne 0\) in \((a,a+\delta)\). Now we replace \(b\) by \(a+\delta.\)

Fix \(x\) in \((a,b)\). For any \(y\) in \((a,x)\), due to Cauchy's mean value theorem, there exists \(\xi\) between \(x\) and \(y\) such that

\[\dfrac{f(x)-f(y)}{g(x)-g(y)}=\dfrac{f'(\xi)}{g'(\xi)}. \]

We rewrite this as

\[\dfrac{f(x)}{g(x)}=\dfrac{f(y)}{g(x)}+\dfrac{f'(\xi)}{g'(\xi)}\left(1-\dfrac{g(y)}{g(x)}\right). \]

If (1), there exists \(\delta_1>0,\delta_2>0\) such that

\[\begin{aligned}y\in (a,a+\delta_1)&\implies |f(y)|<|g(x)|(x-a),\\y\in (a,a+\delta_2)&\implies |g(y)|<|g(x)|(x-a).\end{aligned} \]

Pick \(y_x\in (a,a+\min\{\delta_1,\delta_2,x-a\}).\)

If (2), there exists \(\delta_3>0\) such that

\[y\in (a,a+\delta_3)\implies |g(x)|>\dfrac{\max\{|f(y)|,|g(y)|\}}{x-a} \]

Pick \(y_x\in (a,a+\min\{\delta_3,x-a\})\).

Hence \(\dfrac{f(y_x)}{g(x)},\dfrac{g(y_x)}{g(x)}\rightarrow 0\) as \(x\rightarrow a^{+}\).

Thus there exists \(\xi_x\in (y_x,x)\) such that

\[\dfrac{f(x)}{g(x)}=\dfrac{f(y_x)}{g(x)}+\dfrac{f'(\xi_x)}{g'(\xi_x)}\left(1-\dfrac{g(y_x)}{g(x)}\right). \]

Obviously \(y_x,\xi_x\rightarrow a^{+}\) as \(x\rightarrow a^{+}\), hence

\[\lim_{x\rightarrow a^{+}}\dfrac{f(x)}{g(x)}=\lim_{x\rightarrow a^{+}}\dfrac{f(y_x)}{g(x)}+\lim_{x\rightarrow a^{+}}\dfrac{f'(\xi_x)}{g'(\xi_x)}\left(1-\lim_{x\rightarrow a^{+}}\dfrac{g(y_x)}{g(x)}\right)=\lim_{\xi\rightarrow a^{+}}\dfrac{f'(\xi)}{g'(\xi)}=A. \]