📝 题目
例 12 设 $f\left( x\right)$ 在 $\lbrack 0, + \infty )$ 上一致连续,且对 $\forall h > 0$ ,序列 $\{ f\left( {nh}\right) \}$ 极限存在. 求证: $\mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right)$ 存在.
💡 答案与解析
证 因为 $f\left( x\right)$ 在 $\lbrack 0, + \infty )$ 上一致连续,所以对 $\forall \varepsilon > 0,\exists \delta >$ 0,使得 $\forall \xi ,\eta \geq 0,\left| {\xi - \eta }\right| < \delta$ 时,有
$$ \left| {f\left( \xi \right) - f\left( \eta \right) }\right| < \frac{\varepsilon }{3}. $$
又对上述的 $\varepsilon > 0,\delta > 0$ ,因为 $\mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {n\delta }\right)$ 存
在,所以 $\exists N \in N$ ,使得对 $\forall m,n > N$ ,有
$$ \left| {f\left( {n\delta }\right) - f\left( {m\delta }\right) }\right| < \frac{\varepsilon }{3}.\; < \frac{\varepsilon }{3}\left| \;\right| < \frac{\varepsilon }{3} $$
则对 $\forall {x}_{1},{x}_{2} > X$ ,有
$$ \frac{{x}_{i}}{\delta } > N + 1 \Rightarrow \left\lbrack \frac{{x}_{i}}{\delta }\right\rbrack > N\;\left( {i = 1,2}\right) $$
图 1.5
及
$$ \left| {{x}_{i} - \left\lbrack \frac{{x}_{i}}{\delta }\right\rbrack \delta }\right| = \delta \left| {\frac{{x}_{i}}{\delta } - \left\lbrack \frac{{x}_{i}}{\delta }\right\rbrack }\right| < \delta \;\left( {i = 1,2}\right) . $$
故有 (参见图 1.5).
$$ \left| {f\left( {x}_{1}\right) - f\left( {x}_{2}\right) }\right| = \left| {f\left( {x}_{1}\right) - f\left( {\left\lbrack \frac{{x}_{1}}{\delta }\right\rbrack \delta }\right) }\right| $$
$$ + \left| {f\left( {\left\lbrack \frac{{x}_{2}}{\delta }\right\rbrack \delta }\right) - f\left( {\left\lbrack \frac{{x}_{1}}{\delta }\right\rbrack \delta }\right) }\right| $$
$$ + \left| {f\left( {\left\lbrack \frac{{x}_{2}}{\delta }\right\rbrack \delta }\right) - f\left( {x}_{2}\right) }\right| $$
$$ < \varepsilon < \frac{\varepsilon }{3} + \frac{\varepsilon }{3} + \frac{\varepsilon }{3} = \varepsilon \;\left( {\forall {x}_{1},{x}_{2} > X}\right) . $$
评注 本题采用的证明方法称为 “ $\frac{\varepsilon }{3}$ 论证法”.