Section02_极限重点题型

1. 先和后极限
   • 例 $\displaystyle{\lim_{n\to \infty}\bigg[\frac{1}{1\times 3}+\frac{1}{3\times 5}+\cdots+\frac{1}{(2n-1)(2n+1)}\bigg]}$
   • 解 $$\begin{array}{ll} \because & \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\big(\frac{1}{2n-1}-\frac{1}{2n+1}\big) \\ \therefore & \frac{1}{1\times 3} + \frac{1}{3 \times 5}+\cdots + \frac{1}{(2n-1)(2n+1)}\\ & = \frac{1}{2}(1-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\cdots + \frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})\\ & = \frac{1}{2}(1-\frac{1}{2n+1}) = \frac{2n}{4n+2} \\ \therefore & \lim_{n\to \infty}\big[\cdots\big]=\lim_{n\to\infty}\frac{2n}{4n+2}=\frac{1}{2} \end{array}$$
2. 分子或分母次数不齐 (采用夹逼定理，哪个不齐放大缩小哪个)
   • 例1 $\displaystyle{\lim_{n\to\infty}\bigg(\frac{1}{4n^{2}+1}+\frac{1}{4n^{2}+2}+\cdots+\frac{1}{4n^{2}+n}\bigg)}$
   • 解 $$\begin{array}{ll} & \text{let } \frac{1}{\sqrt{4n^{2}+1}}+\frac{1}{\sqrt{4n^{2}+2}}+\cdots+\frac{1}{\sqrt{4n^{2}+n}}\coloneqq b_n \\ \because & \frac{1}{\sqrt{4n^{2}+n}} \le \frac{1}{\sqrt{4n^{2}+i}} \le \frac{1}{\sqrt{4n^{2}+1}} (1\le i\le n) \\ \therefore & \frac{n}{\sqrt{4n^{2}+n}}\le b_{n}\le \frac{n}{\sqrt{4n^{2}+1}} \\ \because & \lim_{n\to\infty}\frac{n}{\sqrt{4n^{2}+n}} = \frac{1}{2} \\ & \lim_{n\to \infty}\frac{n}{4n^{2}+1} = \frac{1}{2} \\ \therefore & \lim_{n\to \infty}b_{n} = \frac{1}{2} \\ \end{array}$$
   • 例2 $\displaystyle{\lim_{n\to \infty}\bigg(\frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n}{n^{2}+n+n}\bigg)}$
   • 解 $$\begin{array}{ll} & \text{let } \frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n}{n^{2}+n+n}\coloneqq b_{n} \\ \because & \frac{i}{n^{2}+n+n}\le \frac{i}{n^{2}+n+i} \le \frac{i}{n^{2}+n+1}\ (1\le i\le n) \\ \therefore & \sum_{i=1}^{n}\frac{i}{n^{2}+n+n}\le b_{n}\le \sum_{i=1}^{n}\frac{i}{n^{2}+n+1} \\ \because & \lim_{n\to \infty}\sum_{i=1}^{n}\frac{i}{n^{2}+n+n} = \lim_{n\to \infty}\frac{(1+n)n/2}{n^{2}+n+n} = \frac{1}{2} \\ & \lim_{n\to \infty}\sum_{i=1}^{n}\frac{i}{n^{2}+n+1} = \lim_{n\to \infty}\frac{(1+n)n/2}{n^{2}+n+1} = \frac{1}{2} \\ \therefore & \lim_{n\to \infty}b_{n} = \frac{1}{2} \\ \end{array}$$
3. 分子齐次，分母齐次 or 分母比分子多一次 => 采用定积分 $$\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^{n}f\bigg(\frac{i-1}{n}\bigg) = \lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^{n}f\bigg(\frac{i}{n}\bigg) = \int_{0}^{1}f(x)\cdot dx$$
   • 例1 $\displaystyle{\lim_{n\to \infty}\bigg(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\bigg)}$
   • 解 $$\begin{array}{ll} \because & \lim_{n\to \infty}\sum_{i=1}^{n}\frac{1}{n+i} = \lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{i}{n}} \\ & = \int_{0}^{1}\frac{1}{1+x}\cdot dx \\ \therefore & \lim_{n\to \infty}\sum_{i=1}^{n}\frac{1}{n+i} = \int_{0}^{1}\frac{1}{1+x}\cdot dx = \ln(1+x)\vert^{1}_{0} = \ln(2) \\ \end{array}$$
   • 例2 $\displaystyle{\lim_{n\to \infty}\bigg(\frac{n}{n^{2}+1^{2}} + \frac{n}{n^{2}+2^{2}} + \cdots + \frac{n}{n^{2}+n^{2}}\bigg)}$
   • 解 $$\begin{array}{ll} \because & \lim_{n\to \infty}\sum_{i=1}^{n}\frac{n}{n^{2}+i^{2}} = \lim_{n\to \infty}\sum_{i=1}^{n}\frac{1}{n}\frac{1}{1+(\frac{i}{n})^{2}} \\ \therefore & \lim_{n\to \infty}\sum_{i=1}^{n}\frac{n}{n^{2}+i^{2}} = \int_{0}^{1}\frac{1}{1+x^{2}}\cdot dx = \arctan(x)\vert^{1}_{0} = \frac{\pi}{4} \\ \end{array}$$

型二 $\displaystyle{\lim_{n\to\infty}a_{n}\ \exists\ ?}$

准则II：单调有界的数列必有极限

有界

1. 若$\exists\ M>0, \forall\ n, \vert a_{n}\vert<M, \{a_{n}\}$有界
2. 若$\exists\ M_{1}>0, \forall\ n, a_{n}\ge M_{1}, \{a_{n}\}$有下界；若$\exists\ M_{2}>0, \forall\ n, a_{n}\le M_{2}, \{a_{n}\}$有上界
3. $\{a_n\}$有界 $\Leftrightarrow$ $\{a_n\}$ 有上下界

${a_n}$ 单调递增$\begin{cases}\text{无上界 }\Rightarrow\ \lim_{n\to\infty}a_{n}=+\infty\\ a_{n}\le M\ \Rightarrow\ \lim_{n\to\infty}a_{n} \exists\end{cases}$ ${a_n}$ 单调递减$\begin{cases}\text{无下界 }\Rightarrow\ \lim_{n\to\infty}a_{n}=-\infty\\ a_{n}\ge M\ \Rightarrow\ \lim_{n\to\infty}a_{n} \exists\end{cases}$

重要不等式

1. $\color{#D0104C}\text{When }x\ge 0, \sin x\le x$
2. $\color{#D0104C}\text{When }x > 0, \ln{(1+x)} < x$

• 例1：$a_{1}>0$且$a_{n+1}=\ln(a_{n}+1)\quad (n=1,2,\cdots)$，证明$\displaystyle{\lim_{n\to\infty}a_{n}}$存在并求极限
• 解 $$\begin{array}{ll} \because & a_{1}>0 \\ & \text{Suppose } a_{k} > 0\\ \because & a_{k+1} = \ln(a_{k}+1) > 0 \\ \therefore & \forall n,\ \exists a_{n} > 0 \\ \because & a_{n} > 0 \\ \therefore & a_{n+1} = ln(1+a_{n}) < a_{n} \\ \therefore & \{a_{n}\}\text{单调递减} \\ \therefore & \exists\ \lim_{n\to \infty}a_{n} \\\\ & \text{Suppose } \lim_{n\to \infty}a_{n}=A \\ \therefore & A = \ln(1+A) \\ \therefore & A = 0 \\ \therefore & \lim_{n\to \infty}a_{n} = 0 \\ \end{array}$$
• 例2：设 $a_{1}=\sqrt{2},a_{2}=\sqrt{2+\sqrt{2}},a_{3}=\sqrt{2+\sqrt{2+\sqrt{2}}},\cdots$，证明 $\displaystyle{\lim_{n\to\infty}a_{n}}$ 存在并求之
• 解 $$\begin{array}{ll} \because & a_{1} = \sqrt{2}, a_{1}<a_{2},a_{n+1} = \sqrt{2 + a_{n}}\ (n=1,2,\cdots) \\ & \text{Suppose } a_{k} < a_{k+1}\\ \therefore & \sqrt{2+a_{k}} < \sqrt{2 +a_{k+1}} \Rightarrow a_{k+1} < a_{k+2} \\ \therefore & \{a_{n}\}\text{单调递增} \\ \because & a_{1} = \sqrt{2} < 2 \\ & \text{Suppose }a_{k} < 2 \\ \therefore & a_{k+1} = \sqrt{a_{k}+2} < 2 \\ \therefore & \forall\ n, \exists\ a_{n} < 2\\ \therefore & \exists\ \lim_{n\to \infty}a_{n} \\\\ & \text{Suppose } \lim_{n\to \infty}a_{n} = A\\ \therefore & A = \sqrt{2+A} \\ \therefore & A = 2 \\ \therefore & \lim_{n\to \infty}a_{n}=2 \\ \end{array}$$
• 例3：设 $a_{1}=2, a_{n+1} = \frac{1}{2}(a_{n}+\frac{1}{a_{n}})\ (n=1,2,\cdots)$ 证明 $\lim_{n\to\infty}a_{n}$存在
• 解 $$\begin{array}{ll} \because & a_{1} = 2, a_{n+1} = \frac{1}{2}(a_{n}+\frac{1}{a_{n}}) \\ \because & \forall\ x>0, \exists\ x+\frac{1}{x} \ge 2 \\ \therefore & a_{n+1}\ge \frac{1}{2} \times 2 = 1 \\ \therefore & \forall\ n, \exists\ a_{n}\ge 1 \\ \because & a_{n+1} - a_{n} = \frac{1}{2}(-a_{n} + \frac{1}{a_{n}}) \\ \because & a_{n} \ge 1 \\ \therefore & a_{n}\ge \frac{1}{a_{n}} \\ \therefore & a_{n+1}-a_{n} \le 0 \\ & \{a_{n}\}\text{单调递减} \\ \therefore & \exists\ \lim_{n\to \infty}a_{n} \\\\ & \text{Suppose }\lim_{n\to \infty}a_{n} = A \\ \therefore & A = \frac{1}{2}(A+\frac{1}{A}) \\ & \frac{1}{2}A^{2}=\frac{1}{2} \\ \therefore & A = \pm 1 \\ \because & a_{1} = 2 >0 \\ \therefore & \lim_{n\to \infty}a_{n} = 1 \\ \end{array}$$