启发性例题

#1

求：

lim_{x \to + \infty} \sqrt{x} (\sqrt{x + 2} - 2 \sqrt{x + 1} + \sqrt{x})

原式

\begin{aligned} = lim_{x \to + \infty} \sqrt{x} (\sqrt{x + 2} - \sqrt{x + 1}) - \sqrt{x} (\sqrt{x + 1} - \sqrt{x}) \\ = lim_{x \to + \infty} \frac{\sqrt{x}}{(\sqrt{x + 2} + \sqrt{x + 1})} - lim_{x \to + \infty} \frac{\sqrt{x}}{(\sqrt{x + 2} + \sqrt{x + 1})} \\ = lim_{x \to + \infty} \frac{1}{(\sqrt{1 + \frac{2}{x}} + \sqrt{1 + \frac{1}{x}}} - lim_{x \to + \infty} \frac{1}{(\sqrt{1 + \frac{1}{x}} + \sqrt{1}} \\ = \frac{1}{2} - \frac{1}{2} = 0 \end{aligned}

#2

结论：

lim_{x \to x_{0}} u^{v} = e^{lim_{x \to x_{0}} (u - 1) v}

证明：

lim u^{v} = lim {[1 + (u - 1)]^{\frac{1}{u - 1}}}^{(u - 1) v} = e^{lim (u - 1) v}

#3 柯西 - 施瓦尔兹不等式的运用

柯西 - 施瓦尔兹不等式：

{(\sum_{i = 1}^{n} u_{i} v_{i})}^{2} \leq (\sum_{i = 1}^{n} u_{i}^{2}) (\sum_{i = 1}^{n} v_{i}^{2})

已知：

\sum_{k = 1}^{n} a_{k} = 1

求证：

\begin{array}{r} \sum_{k = 1}^{n} (a_{k} + \frac{1}{a_{k}})^{2} \geq \frac{(n^{2} + 1)^{2}}{n} \end{array}

证明：

\begin{aligned} \sum_{k = 1}^{n} (a_{k} + \frac{1}{a_{k}})^{2} \\ = & \frac{1}{n} [\sum_{k = 1}^{n} (a_{k} + \frac{1}{a_{k}})^{2} \times \sum_{k = 1}^{n} 1^{2}] \\ \geq & \frac{1}{n} {[\sum_{k = 1}^{n} (a_{k} + \frac{1}{a_{k}}) \times 1]}^{2} \\ = & \frac{1}{n} {[\sum_{k = 1}^{n} a_{k} + \sum_{k = 1}^{n} a_{k} \times \sum_{k = 1}^{n} \frac{1}{a_{k}}]}^{2} \\ \geq & \frac{1}{n} {1 + {[\sum_{k = 1}^{n} (\sqrt{a_{k}} \times \frac{1}{\sqrt{a_{k}}})]}^{2}}^{2} \\ = & \frac{(n^{2} + 1)^{2}}{n} \end{aligned}

#4

已知 $f (x) = a (x - 1)^{2} + b (x - 1) + c + \sqrt{x^{2} + 3}$ 是 $x$ 趋于 $1$ 时 $(x - 1)^{2}$ 的高阶无穷小，求常数 $a, b, c$ 。

法 1：配凑

设 $t = x - 1, x = t + 1$ 。

\begin{aligned} lim_{x \to 1} \frac{a (x - 1)^{2} + b (x - 1) + c - \sqrt{x^{2} + 3}}{(x - 1)^{2}} = 0 \\ \Rightarrow & lim_{t \to 0} \frac{a t^{2} + b t + c - \sqrt{t^{2} + 2 t + 4}}{t^{2}} = 0 \\ \Rightarrow & lim_{t \to 0} \frac{b t + c - \sqrt{t^{2} + 2 t + 4}}{t^{2}} = - a \\ = & lim_{t \to 0} \frac{(b t + c)^{2} - (t^{2} + 2 t + 4)}{t^{2} (b t + c + \sqrt{t^{2} + 2 t + 4})} \\ = & lim_{t \to 0} \frac{(b^{2} - 1) t^{2} + (2 b c - 2) t + c^{2} - 4}{t^{2} (b t + c + \sqrt{t^{2} + 2 t + 4})} \\ \Rightarrow & {\begin{matrix} 2 b c - 2 = 0 \\ \frac{b^{2} - 1}{c + 2} = - a \\ c^{2} - 4 = 0 \end{matrix} \Rightarrow {\begin{matrix} a = \frac{3}{16} \\ b = \frac{1}{2} \\ c = 2 \end{matrix} \end{aligned}

法 2：洛必达

\begin{aligned} lim_{x \to 1} \frac{a {(x - 1)}^{2} + b (x - 1) + c - \sqrt{x^{2} + 3}}{{(x - 1)}^{2}} = 0 \\ l e t t = x - 1, x = t + 1 \\ lim_{t \to 0} \frac{a t^{2} + b t + c - \sqrt{{(t + 1)}^{2} + 3}}{t^{2}} = lim_{t \to 0} \frac{a t^{2} + b t + c - \sqrt{t^{2} + 2 t + 4}}{t^{2}} = 0 \\ \Rightarrow lim_{t \to 0} a t^{2} + b t + c - \sqrt{t^{2} + 2 t + 4} = c - \sqrt{4} = 0 \Rightarrow c = 2 \\ \Rightarrow lim_{t \to 0} \frac{a t^{2} + b t + 2 - \sqrt{t^{2} + 2 t + 4}}{t^{2}} = a + lim_{t \to 0} \frac{b t + 2 - \sqrt{t^{2} + 2 t + 4}}{t^{2}} = 0 \\ \Rightarrow lim_{t \to 0} \frac{b t + 2 - \sqrt{t^{2} + 2 t + 4}}{t^{2}} = - a \\ lim_{t \to 0} \frac{b t + 2 - \sqrt{t^{2} + 2 t + 4}}{t^{2}} = lim_{t \to 0} \frac{b - \frac{t + 1}{\sqrt{t^{2} + 2 t + 4}}}{2 t} = - a \\ \Rightarrow lim_{t \to 0} b - \frac{t + 1}{\sqrt{t^{2} + 2 t + 4}} = 0 \Rightarrow lim_{t \to 0} \frac{t + 1}{\sqrt{t^{2} + 2 t + 4}} = \frac{1}{\sqrt{4}} = \frac{1}{2} = b \\ \Rightarrow lim_{t \to 0} \frac{b - \frac{t + 1}{\sqrt{t^{2} + 2 t + 4}}}{2 t} = lim_{t \to 0} - \frac{3}{2} (t^{2} + 2 t + 4)^{- \frac{3}{2}} = - \frac{3}{2 \times 2 \times 4} = - \frac{3}{16} = - a \\ \Rightarrow a = \frac{3}{16} \\ \Rightarrow {\begin{matrix} a = \frac{3}{16}, \\ b = \frac{1}{2}, \\ c = 2 \end{matrix} \end{aligned}

注意应用洛必达的条件。

法 3

\begin{aligned} lim_{x \to 1} \frac{a {(x - 1)}^{2} + b (x - 1) + c - \sqrt{x^{2} + 3}}{{(x - 1)}^{2}} = 0 \\ \Rightarrow & lim_{x \to 1} [a (x - 1)^{2} + b (x - 1) + c - \sqrt{x^{2} + 3}] = 0 \\ \Rightarrow & c - \sqrt{4} = 0 \Rightarrow c = 2 \\ lim_{x \to 1} \frac{a (x - 1)^{2} + b (x - 1) + c - \sqrt{x^{2} + 3}}{(x - 1)} = 0 \\ \Rightarrow & b + lim_{x \to 1} \frac{c - \sqrt{x^{2} + 3}}{(x - 1)} \Rightarrow b = \frac{1}{2} \\ lim_{x \to 1} \frac{a {(x - 1)}^{2} + b (x - 1) + c - \sqrt{x^{2} + 3}}{{(x - 1)}^{2}} = 0 \\ \Rightarrow & a + lim_{x \to 1} \frac{\frac{(x - 1)}{2} + 2 - \sqrt{x^{2} + 3}}{{(x - 1)}^{2}} = 0 \Rightarrow a = \frac{3}{16} \\ \Rightarrow & {\begin{matrix} a = \frac{3}{16}, \\ b = \frac{1}{2}, \\ c = 2 \end{matrix} \end{aligned}