三种方法解一道数列极限题

一、题目

$$
\lim_{n \rightarrow \infty} n^{2} \big( \arctan \frac{2}{n} – \arctan \frac{2}{n+1} \big) = ?
$$

难度评级：

二、解析

方法一：等价无穷小 + 取大舍小

$$
\lim_{n \rightarrow \infty} n^{2} \big( \arctan \frac{2}{n} – \arctan \frac{2}{n+1} \big) =
$$

$$
\lim_{n \rightarrow \infty} \big( n^{2} \arctan \frac{2}{n} – n^{2} \arctan \frac{2}{n+1} \big) \Rightarrow
$$

Next

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

$$
\lim_{n \rightarrow \infty} \big( n^{2} \frac{2}{n} – n^{2} \frac{2}{n+1} \big) =
$$

$$
\lim_{n \rightarrow \infty} \big( \frac{2 n^{2}}{n} – \frac{2 n^{2}}{n+1} \big) =
$$

$$
\lim_{n \rightarrow \infty} \frac{2 n^{2} (n+1) – 2n^{3} }{n(n+1)} =
$$

$$
\lim_{n \rightarrow \infty} \frac{2 n^{3} + 2n^{2} – 2n^{3} }{n(n+1)} =
$$

$$
\lim_{n \rightarrow \infty} \frac{ 2n^{2} }{ n^{2} + n } =
$$

$$
\lim_{n \rightarrow \infty} \frac{ 2n^{2} }{ n^{2} } = 2.
$$

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：

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

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

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

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

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

Next

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

$$
\lim_{x \rightarrow 0} \frac{1}{1 + \xi^{2}} = 1.
$$

进而：

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

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

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

$$
\lim_{n \rightarrow \infty} n^{2} \big( \arctan \frac{2}{n} – \arctan \frac{2}{n+1} \big) =
$$

$$
\lim_{x \rightarrow \infty} \frac{2 n^{2}}{n^{2} + n} = \lim_{x \rightarrow \infty} \frac{2 n^{2}}{n^{2}} = 2.
$$

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

$$
\lim_{n \rightarrow \infty} n^{2} \big( \arctan \frac{2}{n} – \arctan \frac{2}{n+1} \big) =
$$

$$
\lim_{n \rightarrow \infty} \frac{\big( \arctan \frac{2}{n} – \arctan \frac{2}{n+1} \big)}{\frac{1}{n^{2}}} =
$$

$$
\lim_{n \rightarrow \infty} \frac{\big[ \arctan \frac{2}{n} – \arctan (\frac{1}{n} \cdot \frac{2}{1 + \frac{1}{n}}) \big]}{\frac{1}{n^{2}}} \Rightarrow
$$

Next

令 $x = \frac{1}{n}$ $\Rightarrow$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{\big[ \arctan 2x – \arctan (x \cdot \frac{2}{1 + x}) \big]}{x^{2}} =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{ \arctan 2x – \arctan \frac{2x}{1 + x} }{x^{2}} \Rightarrow
$$

Next

洛必达法则 $\Rightarrow$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{\frac{2}{1 + 4x^{2}} – \frac{\frac{2(1+x) – 2x}{(1+x)^{2}}}{1 + \frac{4x^{2}}{(1+x)^{2}}} }{2x} =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{\frac{2}{1 + 4x^{2}} – \frac{2(1+x) – 2x}{(1+x)^{2}} \cdot \frac{(1+x)^{2}}{(1+x)^{2} + 4x^{2}} }{2x} =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{\frac{2}{1 + 4x^{2}} – \frac{2(1+x) – 2x}{(1+x)^{2} + 4x^{2}} }{2x} =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{2 [(1+x)^{2} + 4x^{2}] – (1 + 4x^{2})[2(1+x) – 2x] }{(1 + 4x^{2}) [(1+x)^{2} + 4x^{2}] } \cdot \frac{1}{2x} =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{[(1+x)^{2} + 4x^{2}] – (1 + 4x^{2})[(1+x) – x] }{x(1 + 4x^{2}) [(1+x)^{2} + 4x^{2}] } =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{(1+x)^{2} + 4x^{2} – 1 – 4x^{2} }{x(1 + 4x^{2}) [(1+x)^{2} + 4x^{2}] } =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{(1+x)^{2} – 1 }{x(1 + 4x^{2}) [(1+x)^{2} + 4x^{2}] } =
$$

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{ x^{2} + 2x }{x(1 + 4x^{2}) [(1+x)^{2} + 4x^{2}] } =
$$

Next

$$
\lim_{\textcolor{orange}{x \rightarrow 0^{+}} } \frac{ x + 2 }{(1 + 4x^{2}) [(1+x)^{2} + 4x^{2}] } = \frac{2}{1 \cdot 1} = 2.
$$

---