定理1
设函数\(f,g\)在\((a,b)\)上可导，并且\(g(x) \ne 0\)对\(x \in (a,b)\)成立。又设
\[ \lim \limits_{x \to a^+} f(x) = \lim \limits_{x \to a^+} g(x) = 0 \]
如果极限
\[ \lim \limits_{x \to a^+} \frac{f^\prime(x)}{g^\prime(x)} \]
存在(或为\(\infty\))，那么便有
\[ \lim \limits_{x \to a^+} \frac{f(x)}{g(x)} = \lim \limits_{x \to a^+} \frac{f^\prime(x)}{g^\prime(x)} \]
证：补充定义\(f(a) = g(a) =
0\)，从而\(f,g\)在\([a,b)\)上连续，利用Cauchy中值定理，对\(x \in (a,b)\)，有
\[
\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} =
\frac{f^\prime(\xi)}{g^\prime(\xi)}
\]
这里\(a < \xi < x\)，所以当\(x \to a^+\)时，有\(\xi \to a^+\)，从而
\[
\lim\limits_{x \to a^+} \frac{f(x)}{g(x)} = \lim\limits_{\xi \to
a^+} \frac{f^\prime(\xi)}{g^\prime(\xi)} = \lim\limits_{x \to a^+}
\frac{f^\prime(x)}{g^\prime(x)}
\]
Q.E.D.
定理2
设函数\(f,g\)在区间\((a,+\infty)\)上可导，且\(g(x) \ne 0\)对\(x \in (a, +\infty)\)成立，并且
\[ \lim\limits_{x \to +\infty} f(x) = \lim\limits_{x \to +\infty} g(x) = 0 \]
那么当\(\lim \limits_{x \to +\infty} \dfrac{f^\prime(x)}{g^\prime(x)}\)存在（或为\(\infty\)）时，有
\[ \lim \limits_{x \to +\infty} \frac{f(x)}{g(x)} = \lim \limits_{x \to +\infty} \frac{f^\prime(x)}{g^\prime(x)} \]
证：令\(x =
\dfrac{1}{t}\)，则当\(x \to
+\infty\)相当于\(t \to
0^+\)，这时，我们有
\[
\lim\limits_{t \to 0^+} f\left(\frac{1}{t}\right) = \lim\limits_{t
\to 0^+} g\left(\frac{1}{t}\right) = 0
\]
由定理1可知
\[
\begin{aligned}
\lim\limits_{x \to +\infty} \frac{f(x)}{g(x)} = \lim\limits_{t \to
0^+} \frac{f\left(\frac{1}{t}\right)}{g\left(\frac{1}{t}\right)} =
\frac{f^\prime\left(\frac{1}{t}\right)\left(-\frac{1}{t^2}\right)}{g^\prime\left(\frac{1}{t}\right)\left(-\frac{1}{t^2}\right)}
\\ = \lim\limits_{t \to 0^+}
\frac{f^\prime\left(\frac{1}{t}\right)}{g^\prime\left(\frac{1}{t}\right)}
= \lim\limits_{x \to +\infty} \frac{f^\prime(x)}{g^\prime(x)}
\end{aligned}
\]
Q.E.D.
定理3
设函数\(f,g\)在\((a,b)\)上可导，\(g(x) \ne 0\)，且
\[ \lim\limits_{x \to a^+} g(x) = \infty \]
如果极限\(\lim \limits_{x \to a^+} \frac{f^\prime(x)}{g^\prime(x)}\)存在（或为\(\infty\)），那么
\[ \lim \limits_{x \to a^+} \frac{f(x)}{g(x)} = \lim \limits_{x \to a^+} \frac{f^\prime(x)}{g^\prime(x)} \]
证：令
\[
l = \lim\limits_{x \to a^+} \frac{g^\prime(x)}{g^\prime(x)}
\]
不妨设\(l\)为有限数，则对任意的\(\varepsilon > 0\)，存在\(\delta > 0\)，使得当\(x \in (a, a+\delta)\)时，有
\[
l - \varepsilon < \frac{f^\prime(x)}{g^\prime(x)} < l +
\varepsilon
\]
从而对\((x,c) \in (a,
a+\delta)\)，由Cauchy中值定理可知，必存在\(\xi \in (x,c)\)使得
\[
l - \varepsilon < \frac{f(x) - f(c)}{g(x) - g(c)} =
\frac{f^\prime(\xi)}{g^\prime(\xi)} < l + \varepsilon
\]
又因为
\[
\frac{f(x)-f(c)}{g(x) - g(c)} = \left( \frac{f(x)}{g(x)} -
\frac{f(c)}{g(x)}\right) \left(1 - \frac{g(c)}{g(x)}\right)^{-1}
\]
固定\(c\)，对\(x \to a^+\)取上极限，得
\[
\limsup_{x \to a^+} \frac{f(x)}{g(x)} \le l + \varepsilon
\]
再令\(\varepsilon \to 0\)，得
\[
\limsup_{x \to a^+} \frac{f(x)}{g(x)} \le l
\]
利用同样得方式，可得
\[
\liminf_{x \to a^+} \frac{f(x)}{g(x)} \ge l
\]
从而有
\[
\lim\limits_{x \to a^+} \frac{f(x)}{g(x)} = l
\]
当\(l\)为\(-\infty\)或\(+\infty\)也可使用类似得方法证明。
Q.E.D.