函数 $f\left( x \right)$ 的导函数 $f'\left( x \right)$ 连续，且 $f\left( 0 \right) = 0$，$f'\left( 0 \right) = a$．记曲线 $y = f\left( x \right)$ 与 $P\left(t,0\right)$ 最近的点为 $Q\left(s,f(s)\right)$，求极限值 $\mathop {\lim }\limits_{t \to 0} \dfrac{s}{t}$．
【难度】
【出处】
2006年北京大学自主招生保送生测试
【标注】
【答案】
$\dfrac{1}{{{a^2} + 1}}$
【解析】
函数在 $Q$ 点处的切线必须与以 $P$ 为圆心，$PQ$ 为半径的圆相切，也即与$${\left( {x - t} \right)^2} + {y^2} = {\left( {s - t} \right)^2} + {\left[ {f\left( s \right)} \right]^2}$$相切．
于是切线为$$\left( {s - t} \right)\left( {x - t} \right) + f\left( s \right) \cdot y = {\left( {s - t} \right)^2} + {\left[ {f\left( s \right)} \right]^2}.$$所以$$f'\left( s \right) = \frac{{t - s}}{{f\left( s \right)}},$$整理得$$\dfrac{s}{t} = 1 - \dfrac{{f\left( s \right)f'\left( s \right)}}{t}.$$所以$$ \lim\limits_{t \to 0} \frac{s}{t} =\lim\limits_{t \to 0} \left[ {1 - \frac{{f\left( s \right)f'\left( s \right)}}{t}} \right] = 1 -\lim\limits_{t \to 0} f'\left( s \right) \cdot \mathop {\lim }\limits_{t \to 0} \frac{{f\left( s \right)}}{s} \cdot \mathop {\lim }\limits_{t \to 0} \frac{s}{t} = 1 - {a^2} \cdot \lim\limits_{t \to 0} \frac{s}{t}.$$于是$$\lim \limits_{t \to 0} \frac{s}{t} = \frac{1}{{{a^2} + 1}}.$$
于是切线为$$\left( {s - t} \right)\left( {x - t} \right) + f\left( s \right) \cdot y = {\left( {s - t} \right)^2} + {\left[ {f\left( s \right)} \right]^2}.$$所以$$f'\left( s \right) = \frac{{t - s}}{{f\left( s \right)}},$$整理得$$\dfrac{s}{t} = 1 - \dfrac{{f\left( s \right)f'\left( s \right)}}{t}.$$所以$$ \lim\limits_{t \to 0} \frac{s}{t} =\lim\limits_{t \to 0} \left[ {1 - \frac{{f\left( s \right)f'\left( s \right)}}{t}} \right] = 1 -\lim\limits_{t \to 0} f'\left( s \right) \cdot \mathop {\lim }\limits_{t \to 0} \frac{{f\left( s \right)}}{s} \cdot \mathop {\lim }\limits_{t \to 0} \frac{s}{t} = 1 - {a^2} \cdot \lim\limits_{t \to 0} \frac{s}{t}.$$于是$$\lim \limits_{t \to 0} \frac{s}{t} = \frac{1}{{{a^2} + 1}}.$$
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