准则Ⅰ如果数列{xn},{yn}及{zn}满足下列条件:
(1)从某项起,即∃n0∈N,当n>n0 时,有yn≤xn≤zn,
(2)limn→∞yn=a,limn→∞zn=a,
那么数列{xn}的极限存在,且limn→∞xn=a.
证明:因yn→a,zn→a,所以根据数列极限的定义,∀ε>0,\exists 正整数N_1,当n>N_1时,有|y_n-a|<ε;又存在正整数N_2,当n>N_2时,有|z_n-a|<ε.取N = \max \{ {n_0},{N_1},{N_2}\} ,则当n>N时,有|y_n-a|<ε,|z_n-a|<ε同时成立,即
a - \varepsilon < {y_n} < a + \varepsilon ,a - \varepsilon < {z_n} < a + \varepsilon ,且y_n≤x_n≤z_n,则
a - \varepsilon < {y_n} \leqslant {x_n} \leqslant {z_n} < a + \varepsilon ,即|x_n-a|<ε
这就证明了\mathop {\lim }\limits_{n \to \infty } {x_n} = a.
数列极限的存在准则可以推广到函数的极限:
准则Ⅰ'如果
(1)当x∈\mathop U\limits^0 \left( {x_0,r } \right)(或|x|>M)时,
g(x) \leqslant f(x) \leqslant h(x),(2) \mathop {\lim }\limits_{x \to {x_0}} g(x) = A,\mathop {\lim }\limits_{x \to {x_0}} h(x) = A,或\mathop {\lim }\limits_{x \to \infty } g(x) = A,\mathop {\lim }\limits_{x \to \infty } h(x) = A,
那么\mathop {\lim }\limits_{x \to {x_0}} f(x) = A或\mathop {\lim }\limits_{x \to \infty } f(x) = A.
例 证明
\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1.证明 首先注意到函数f(x) = \frac{{\sin x}}{x}的定义域是:\{ x|x \in {\bf{R}}但x \ne 0\} ,而且证明的是x→0(因为函数在x=0处没有定义,而通常情况下我们想知道函数在没有定义或无法取值的点的函数值)时函数的极限.在一维的情况下, x→0只包括x从0的左侧和右侧同时趋近,且\sin x为三角函数,因此我们只需考虑0的某一去心邻域\mathop U\limits^0 \left( {0,\frac{\pi }{2}} \right)即可,即x \in \mathop U\limits^0 \left( {0,{\pi \over 2}} \right).
\frac{{\sin x}}{x}在x=0时没有意义,是我们求证x→0时\frac{{\sin x}}{x}的极限的原因.
如图所示,\angle DOA = x(0 < x < \frac{\pi }{2})
OC = \cos x,CB = \sin x,AD = \tan x,\widehat {AB} = x,
{S_{\Delta OAB}} < 扇形OAB面积 < {S_{\Delta OAD}},
\frac{{\pi {R^2}}}{{2\pi }} = \frac{{扇形OAB面积}}{x} \Rightarrow 扇形OAB面积 = \frac{x}{2},所以 \begin{equation} \frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x \end{equation}
或 \begin{equation} \cos x < \frac{{\sin x}}{x} < 1 \end{equation}
在第四象限内依然比较面积关系,有
\frac{{ - \tan ( - x)}}{2} > \frac{{ - ( - x)}}{2} > \frac{{ - \sin ( - x)}}{2} > 0, - x < \sin ( - x) < 0, \left\{ \begin{gathered} 0 < \sin x < x, \hfill \\ - x < \sin ( - x) < 0, \hfill \\ \end{gathered} \right. \Leftrightarrow |x| > |\sin x|,x \in \mathop U\limits^0 \left( {0,{\pi \over 2}} \right), \frac{{ - \tan ( - x)}}{2} > \frac{{ - ( - x)}}{2} > \frac{{ - \sin ( - x)}}{2} > 0, \tan ( - x) < - x < \sin ( - x) < 0, \frac{1}{{\cos ( - x)}} > \frac{{ - x}}{{\sin ( - x)}} > 1, \cos ( - x) < \frac{{\sin ( - x)}}{{ - x}} < 1,这与(2)式是一致的,
因此用-x代替x时,(2)式对于开区间( - \frac{\pi }{2},0)内的一切点x也是成立的.
当|x| \geqslant \frac{\pi }{2}时,\sin x| \leqslant 1 < |x|,因此对于任意的角度x,当x≠0 时有|\sin x| < |x|.
为了对上式运用准则Ⅰ',下面来证明
\mathop {\lim }\limits_{x \to 0} \cos x = 1.当0 < |x| < \frac{\pi }{2}时,
0 < |\cos x - 1| = 1 - \cos x = 2{\sin ^2}\frac{x}{2}因为|x| > |\sin x|,x \in \mathop U\limits^0 \left( {0,{\pi \over 2}} \right),因此
0 < |\cos x - 1| = 1 - \cos x = 2{\sin ^2}\frac{x}{2} < 2{(\frac{x}{2})^2} = \frac{{{x^2}}}{2},当x \to 0,\frac{{{x^2}}}{2} \to 0,由准则Ⅰ'有\mathop {\lim }\limits_{x \to 0} (1 - \cos x) = 0,由函数极限的运算法则,有\mathop {\lim }\limits_{x \to 0} \cos x = 1,由于\mathop {\lim }\limits_{x \to 0} \cos x = 1,\mathop {\lim }\limits_{x \to 0} 1 = 1,由不等式 (2) 及准则Ⅰ',即得
\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1.