三种方法解一道数列极限题

一、题目

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2(\arctan \frac{2}{n}–\arctan \frac{2}{n+1})=?$$

难度评级：

二、解析

方法一：等价无穷小 + 取大舍小

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2(\arctan \frac{2}{n}–\arctan \frac{2}{n+1})=$$

$$\underset{n\to \mathrm{\infty }}{lim}(n^2\arctan \frac{2}{n}–n^2\arctan \frac{2}{n+1})\Rightarrow$$

Next

有等价无穷小 $\arctan x$ $\sim$ $\tan x$ $\sim$ $\arcsin x$ $\sim$ $\sin x$ $\sim$ $x$, 于是 $\Rightarrow$

$$\underset{n\to \mathrm{\infty }}{lim}(n^2\frac{2}{n}–n^2\frac{2}{n+1})=$$

$$\underset{n\to \mathrm{\infty }}{lim}(\frac{2n^2}{n}–\frac{2n^2}{n+1})=$$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{2n^2(n+1)–2n^3}{n(n+1)}=$$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{2n^3+2n^2–2n^3}{n(n+1)}=$$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{2n^2}{n^2+n}=$$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{2n^2}{n^2}=2.$$

方法二：拉格朗日中值定理

$\arctan \frac{2}{n}$ 和 $\arctan \frac{2}{n+1}$ 可以看作是函数 $f(x)$ $=$ $\arctan x$ 取不同变量所形成的式子。

因此，根据拉格朗日中值定理，有：

$$\frac{f(\frac{2}{n})–f(\frac{2}{n+1})}{\frac{2}{n}–\frac{2}{n+1}}=f^{\prime }(\xi ).$$

其中，$\xi$ $\in$ $(\frac{2}{n},\frac{2}{n+1})$.

Next

于是：

$$\frac{f(\frac{2}{n})–f(\frac{2}{n+1})}{\frac{2}{n}–\frac{2}{n+1}}=f^{\prime }(\xi )\Rightarrow$$

$$\frac{f(\frac{2}{n})–f(\frac{2}{n+1})}{\frac{2}{n}–\frac{2}{n+1}}=(\arctan \xi {)}^{\prime }\Rightarrow$$

$$\frac{f(\frac{2}{n})–f(\frac{2}{n+1})}{\frac{2}{n}–\frac{2}{n+1}}=\frac{1}{1+{\xi }^2}\Rightarrow$$

$$f(\frac{2}{n})–f(\frac{2}{n+1})=\frac{1}{1+{\xi }^2}(\frac{2}{n}–\frac{2}{n+1})\Rightarrow$$

$$f(\frac{2}{n})–f(\frac{2}{n+1})=\frac{1}{1+{\xi }^2}\cdot \frac{2}{n^2+n}.$$

Next

由于 $\xi$ $\in$ $(\frac{2}{n},\frac{2}{n+1})$, 于是，当 $n\to \mathrm{\infty }$ 时，$\xi \to 0$, 即：

$$\underset{x\to 0}{lim}\frac{1}{1+{\xi }^2}=1.$$

进而：

$$f(\frac{2}{n})–f(\frac{2}{n+1})=\frac{1}{1+{\xi }^2}\cdot \frac{2}{n^2+n}\Rightarrow$$

$$f(\frac{2}{n})–f(\frac{2}{n+1})=\frac{2}{n^2+n}\Rightarrow$$

$$\arctan \frac{2}{n}–\arctan \frac{2}{n+1}=\frac{2}{n^2+n}\Rightarrow$$

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2(\arctan \frac{2}{n}–\arctan \frac{2}{n+1})=$$

$$\underset{x\to \mathrm{\infty }}{lim}\frac{2n^2}{n^2+n}=\underset{x\to \mathrm{\infty }}{lim}\frac{2n^2}{n^2}=2.$$

方法三：将 $\mathrm{\infty }\cdot 0$ 型转为 $0\cdot 0$ 型并用洛必达法则直接算

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2(\arctan \frac{2}{n}–\arctan \frac{2}{n+1})=$$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{(\arctan \frac{2}{n}–\arctan \frac{2}{n+1})}{\frac{1}{n^2}}=$$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{[\arctan \frac{2}{n}–\arctan (\frac{1}{n}\cdot \frac{2}{1+\frac{1}{n}})]}{\frac{1}{n^2}}\Rightarrow$$

Next

令 $x=\frac{1}{n}$ $\Rightarrow$

$$\underset{x\to 0^{+}}{lim}\frac{[\arctan 2x–\arctan (x\cdot \frac{2}{1+x})]}{x^2}=$$

$$\underset{x\to 0^{+}}{lim}\frac{\arctan 2x–\arctan \frac{2x}{1+x}}{x^2}\Rightarrow$$

Next

洛必达法则 $\Rightarrow$

$$\underset{x\to 0^{+}}{lim}\frac{\frac{2}{1+4x^2}–\frac{\frac{2(1+x)–2x}{(1+x{)}^2}}{1+\frac{4x^2}{(1+x{)}^2}}}{2x}=$$

$$\underset{x\to 0^{+}}{lim}\frac{\frac{2}{1+4x^2}–\frac{2(1+x)–2x}{(1+x{)}^2}\cdot \frac{(1+x{)}^2}{(1+x{)}^2+4x^2}}{2x}=$$

$$\underset{x\to 0^{+}}{lim}\frac{\frac{2}{1+4x^2}–\frac{2(1+x)–2x}{(1+x{)}^2+4x^2}}{2x}=$$

$$\underset{x\to 0^{+}}{lim}\frac{2[(1+x{)}^2+4x^2]–(1+4x^2)[2(1+x)–2x]}{(1+4x^2)[(1+x{)}^2+4x^2]}\cdot \frac{1}{2x}=$$

$$\underset{x\to 0^{+}}{lim}\frac{[(1+x{)}^2+4x^2]–(1+4x^2)[(1+x)–x]}{x(1+4x^2)[(1+x{)}^2+4x^2]}=$$

$$\underset{x\to 0^{+}}{lim}\frac{(1+x{)}^2+4x^2–1–4x^2}{x(1+4x^2)[(1+x{)}^2+4x^2]}=$$

$$\underset{x\to 0^{+}}{lim}\frac{(1+x{)}^2–1}{x(1+4x^2)[(1+x{)}^2+4x^2]}=$$

$$\underset{x\to 0^{+}}{lim}\frac{x^2+2x}{x(1+4x^2)[(1+x{)}^2+4x^2]}=$$

Next

$$\underset{x\to 0^{+}}{lim}\frac{x+2}{(1+4x^2)[(1+x{)}^2+4x^2]}=\frac{2}{1\cdot 1}=2.$$

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