启发性例题

#1

求：

$$\underset{x\to +\infty }{\lim }\sqrt{x}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$$

原式

$$\begin{array}{rl}& =\underset{x\to +\infty }{\lim }\sqrt{x}(\sqrt{x+2}-\sqrt{x+1})-\sqrt{x}(\sqrt{x+1}-\sqrt{x})\\ & =\underset{x\to +\infty }{\lim }\frac{\sqrt{x}}{(\sqrt{x+2}+\sqrt{x+1})}-\underset{x\to +\infty }{\lim }\frac{\sqrt{x}}{(\sqrt{x+2}+\sqrt{x+1})}\\ & =\underset{x\to +\infty }{\lim }\frac{1}{(\sqrt{1+\frac{2}{x}}+\sqrt{1+\frac{1}{x}}}-\underset{x\to +\infty }{\lim }\frac{1}{(\sqrt{1+\frac{1}{x}}+\sqrt{1}}\\ & =\frac{1}{2}-\frac{1}{2}=0\end{array}$$

#2

结论：

$$\underset{x\to x_0}{\lim }{u}^v=e^{\underset{x\to x_0}{\lim }(u-1)v}$$

证明：

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

#3 柯西 - 施瓦尔兹不等式的运用

柯西 - 施瓦尔兹不等式：

$${\left(\sum _{i=1}^n{u}_i{v}_i\right)}^2\le \left(\sum _{i=1}^n{u}_i^2\right)\left(\sum _{i=1}^n{v}_i^2\right)$$

已知：

$$\sum _{k=1}^n{a}_k=1$$

求证：

$$\begin{array}{r}\sum _{k=1}^n(a_k+\frac{1}{a_k}{)}^2\ge \frac{(n^2+1{)}^2}{n}\end{array}$$

证明：

$$\begin{array}{rl}& \sum _{k=1}^n(a_k+\frac{1}{a_k}{)}^2\\ =& \frac{1}{n}\left[\sum _{k=1}^n(a_k+\frac{1}{a_k}{)}^2\times \sum _{k=1}^n{1}^2\right]\\ \ge & \frac{1}{n}{\left[\sum _{k=1}^n(a_k+\frac{1}{a_k})\times 1\right]}^2\\ =& \frac{1}{n}{\left[\sum _{k=1}^n{a}_k+\sum _{k=1}^n{a}_k\times \sum _{k=1}^n\frac{1}{a_k}\right]}^2\\ \ge & \frac{1}{n}{\left\{1+{\left[\sum _{k=1}^n(\sqrt{a_k}\times \frac{1}{\sqrt{a_k}})\right]}^2\right\}}^2\\ =& \frac{(n^2+1{)}^2}{n}\end{array}$$

#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 $t=x-1,x=t+1$。

$$\begin{array}{rl}& \underset{x\to 1}{\lim }\frac{a(x-1{)}^2+b(x-1)+c-\sqrt{x^2+3}}{(x-1{)}^2}=0\\ \Rightarrow & \underset{t\to 0}{\lim }\frac{a{t}^2+bt+c-\sqrt{t^2+2t+4}}{t^2}=0\\ \Rightarrow & \underset{t\to 0}{\lim }\frac{bt+c-\sqrt{t^2+2t+4}}{t^2}=-a\\ =& \underset{t\to 0}{\lim }\frac{(bt+c{)}^2-(t^2+2t+4)}{t^2(bt+c+\sqrt{t^2+2t+4})}\\ =& \underset{t\to 0}{\lim }\frac{(b^2-1)t^2+(2bc-2)t+c^2-4}{t^2(bt+c+\sqrt{t^2+2t+4})}\\ \Rightarrow & \{\begin{array}{l}2bc-2=0\\ \frac{b^2-1}{c+2}=-a\\ c^2-4=0\end{array}\Rightarrow \{\begin{array}{l}a=\frac{3}{16}\\ b=\frac{1}{2}\\ c=2\end{array}\end{array}$$

法 2：洛必达

$$\begin{array}{rl}& \underset{x\to 1}{\lim }\frac{a{(x-1)}^2+b(x-1)+c-\sqrt{x^2+3}}{{(x-1)}^2}=0\\ & \mathrm{let}t=x-1,x=t+1\\ & \underset{t\to 0}{\lim }\frac{a{t}^2+bt+c-\sqrt{{(t+1)}^2+3}}{t^2}=\underset{t\to 0}{\lim }\frac{a{t}^2+bt+c-\sqrt{t^2+2t+4}}{t^2}=0\\ & \Rightarrow \underset{t\to 0}{\lim }a{t}^2+bt+c-\sqrt{t^2+2t+4}=c-\sqrt{4}=0\Rightarrow c=2\\ & \Rightarrow \underset{t\to 0}{\lim }\frac{a{t}^2+bt+2-\sqrt{t^2+2t+4}}{t^2}=a+\underset{t\to 0}{\lim }\frac{bt+2-\sqrt{t^2+2t+4}}{t^2}=0\\ & \Rightarrow \underset{t\to 0}{\lim }\frac{bt+2-\sqrt{t^2+2t+4}}{t^2}=-a\\ & \underset{t\to 0}{\lim }\frac{bt+2-\sqrt{t^2+2t+4}}{t^2}=\underset{t\to 0}{\lim }\frac{b-\frac{t+1}{\sqrt{t^2+2t+4}}}{2t}=-a\\ & \Rightarrow \underset{t\to 0}{\lim }b-\frac{t+1}{\sqrt{t^2+2t+4}}=0\Rightarrow \underset{t\to 0}{\lim }\frac{t+1}{\sqrt{t^2+2t+4}}=\frac{1}{\sqrt{4}}=\frac{1}{2}=b\\ & \Rightarrow \underset{t\to 0}{\lim }\frac{b-\frac{t+1}{\sqrt{t^2+2t+4}}}{2t}=\underset{t\to 0}{\lim }-\frac{3}{2}(t^2+2t+4{)}^{-\frac{3}{2}}=-\frac{3}{2\times 2\times 4}=-\frac{3}{16}=-a\\ & \Rightarrow a=\frac{3}{16}\\ & \Rightarrow \{\begin{array}{l}a=\frac{3}{16},\\ b=\frac{1}{2},\\ c=2\end{array}\end{array}$$

注意应用洛必达的条件。

法 3

$$\begin{array}{rl}& \underset{x\to 1}{\lim }\frac{a{(x-1)}^2+b(x-1)+c-\sqrt{x^2+3}}{{(x-1)}^2}=0\\ \Rightarrow & \underset{x\to 1}{\lim }[a(x-1{)}^2+b(x-1)+c-\sqrt{x^2+3}]=0\\ \Rightarrow & c-\sqrt{4}=0\Rightarrow c=2\\ & \underset{x\to 1}{\lim }\frac{a(x-1{)}^2+b(x-1)+c-\sqrt{x^2+3}}{(x-1)}=0\\ \Rightarrow & b+\underset{x\to 1}{\lim }\frac{c-\sqrt{x^2+3}}{(x-1)}\Rightarrow b=\frac{1}{2}\\ & \underset{x\to 1}{\lim }\frac{a{(x-1)}^2+b(x-1)+c-\sqrt{x^2+3}}{{(x-1)}^2}=0\\ \Rightarrow & a+\underset{x\to 1}{\lim }\frac{\frac{(x-1)}{2}+2-\sqrt{x^2+3}}{{(x-1)}^2}=0\Rightarrow a=\frac{3}{16}\\ \Rightarrow & \{\begin{array}{l}a=\frac{3}{16},\\ b=\frac{1}{2},\\ c=2\end{array}\end{array}$$