设
lim
x→a+f(x)=limx→+∞f(x)=A <script id="MathJax-Element-1" type="math/tex; mode=display">\mathop {\lim }\limits_{x \to \begin{array}{*{20}{c}} {{a^ + }} \end{array}} f\left( x \right) = \mathop {\lim }\limits_{x \to \begin{array}{*{20}{c}} { + \infty } \end{array}} f\left( x \right) = A</script>其中$A$是有限数或$\pm \infty $
若$f\left( x \right) = A$,则结论显然成立；若$f\left( x \right) \ne A$,则存在${x_0} \in \left( {a, + \infty } \right)$,使得$f\left( {{x_0}} \right) \ne A$.
不妨设$f\left( {{x_0}} \right) > A$,则由实数的稠密性知,存在${\varepsilon _0} > 0$,使得
<script id="MathJax-Element-2" type="math/tex; mode=display">f\left( {{x_0}} \right) > f\left( {{x_0}} \right) - {\varepsilon _0} > A</script>由$\lim \limits_{x \to
<script id="MathJax-Element-3" type="math/tex; mode=display">\begin{array}{*{20}{c}} {{a^ + }} \end{array}</script>} f\left( x \right) = A < A + {\varepsilon _0}$及极限的保号性知
?δ>0,?x∈(a,a+δ),有f(x)<A+ε0 <script id="MathJax-Element-4" type="math/tex; mode=display">\exists \delta > 0,\forall x \in \left( {a,a + \delta } \right),有f\left( x \right) < A + {\varepsilon _0}</script>特别地,取${x_1} \in \left( {a,a + \delta } \right)$,且${x_1} < {x_0}$,则
<script id="MathJax-Element-5" type="math/tex; mode=display">f\left( {{x_1}} \right) < A + {\varepsilon _0} < f\left( {{x_0}} \right)</script>由连续函数介值定理知,存在${\xi _1} \in \left( {{x_1},{x_0}} \right)$,使得
<script id="MathJax-Element-6" type="math/tex; mode=display">f\left( {{\xi _1}} \right) = A + {\varepsilon _0}</script>由$\lim \limits_{x \to
<script id="MathJax-Element-7" type="math/tex; mode=display">\begin{array}{*{20}{c}} { + \infty } \end{array}</script>} f\left( x \right) = A < A + {\varepsilon _0}$及极限的保号性知
<script id="MathJax-Element-8" type="math/tex; mode=display">\exists M > a,\forall x > M,有f\left( x \right) < A + {\varepsilon _0}</script>
特别地,取${x_2} \in \left( {M, + \infty } \right)$,且${x_0} < {x_2}$,则
<script id="MathJax-Element-9" type="math/tex; mode=display">f\left( {{x_2}} \right) < A + {\varepsilon _0} < f\left( {{x_0}} \right)</script>由连续函数介值定理知,存在${\xi _2} \in \left( {{x_0},{x_2}} \right)$,使得
<script id="MathJax-Element-10" type="math/tex; mode=display">f\left( {{\xi _2}} \right) = A + {\varepsilon _0}</script>由$Rolle$中值定理知,存在$\xi \in \left( {{\xi _1},{\xi _2}} \right)$,使得
<script id="MathJax-Element-11" type="math/tex; mode=display">f‘\left( \xi \right) = 0</script>
