📝 题目
例 29 已知 $f\left( x\right)$ 在 $\lbrack 0, + \infty )$ 上有二阶连续导数, $f\left( 0\right) = {f}^{\prime }\left( 0\right)$ $= 0$ ,且 ${f}^{\prime \prime }\left( x\right) > 0$ . 若对任意的 $x > 0$ ,函数 $u\left( x\right)$ 表示曲线 $y = f\left( x\right)$ 在切点 $\left( {x,f\left( x\right) }\right)$ 处的切线在 $x$ 轴上的截距 (如图 3.5 所示).
(1) 写出 $u\left( x\right)$ 的表达式,并求 $\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}u\left( x\right)$ 及 $\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}{u}^{\prime }\left( x\right)$ ;
(2) 求 $\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{{\int }_{0}^{u\left( x\right) }f\left( t\right) \mathrm{d}t}{{\int }_{0}^{x}f\left( t\right) \mathrm{d}t}$ .
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图 3.5
💡 答案与解析
解 (1) $u\left( x\right) = x - \frac{f\left( x\right) }{{f}^{\prime }\left( x\right) }$ ,
$$ \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}u\left( x\right) = - \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{f\left( x\right) }{{f}^{\prime }\left( x\right) } = \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{{f}^{\prime }\left( x\right) }{{f}^{\prime \prime }\left( x\right) } = 0; $$
$$ {u}^{\prime }\left( x\right) = 1 - \frac{{f}^{\prime }{\left( x\right) }^{2} - {f}^{\prime \prime }\left( x\right) f\left( x\right) }{{f}^{\prime }{\left( x\right) }^{2}} = \frac{{f}^{\prime \prime }\left( x\right) f\left( x\right) }{{f}^{\prime }{\left( x\right) }^{2}}, $$
$$ \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}{u}^{\prime }\left( x\right) = {f}^{\prime \prime }\left( 0\right) \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{{f}^{\prime }\left( x\right) }{2{f}^{\prime }\left( x\right) {f}^{\prime \prime }\left( x\right) } = \frac{1}{2}. $$
(2) $\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{{\int }_{0}^{u\left( x\right) }f\left( t\right) \mathrm{d}t}{{\int }_{0}^{x}f\left( t\right) \mathrm{d}t} = \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{f\left( {u\left( x\right) }\right) {u}^{\prime }\left( x\right) }{f\left( x\right) } = \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{f\left( {u\left( x\right) }\right) }{f\left( x\right) }$
$$ = \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{{f}^{\prime }\left( {u\left( x\right) }\right) {u}^{\prime }\left( x\right) }{{f}^{\prime }\left( x\right) } $$
$$ = \frac{1}{4}\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{{f}^{\prime \prime }\left( {u\left( x\right) }\right) {u}^{\prime }\left( x\right) }{{f}^{\prime \prime }\left( x\right) } = \frac{1}{8}. $$