2021 数学分析

1. 求极限

   $$\underset{n\to \infty }{\lim }\frac{1}{n}\sqrt[n]{(n+1)(n+2)\cdots (n+n)}$$

$$\begin{array}{rl}\underset{n\to \infty }{\lim }\frac{1}{n}\sqrt[n]{(n+1)(n+2)\cdots (n+n)}& =\underset{n\to \infty }{\lim }\sqrt[n]{\frac{(n+1)(n+2)\cdots (n+n)}{n^m}}\\ & =\underset{n\to \infty }{\lim }{e}^{\frac{1}{n}\sum _{k=1}^n\ln \left(1+\frac{k}{n}\right)}\\ & =e^{{\int }_0^1\ln \left(1+x\right)dx}\\ & =e^{2ln2-1}\end{array}$$

2. 求 $a,b$ 的值，使得

   $$\underset{x\to 0}{\lim }\frac{1}{bx-\sin x}{\int }_0^x\frac{t^2}{\sqrt{a+t^2}}dt=1.$$

$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{{\int }_0^x\frac{t^2}{\sqrt{a+t^2}}dt}{bx-\sin x}& =\underset{x\to 0}{\lim }\frac{x^2}{\left(b-\cos x\right)\sqrt{a+x^2}}\\ & =\underset{x\to 0}{\lim }\frac{x^2}{\left(b-1+\frac{1}{2}{x}^2+o(x^2)\right)\left(\sqrt{a}+\frac{1}{2\sqrt{a}}{x}^2+o(x^2)\right)}\\ & =\underset{x\to 0}{\lim }\frac{x^2}{\left(\left(b-1\right)\sqrt{a}+\frac{1}{2}\left(\sqrt{a}+\frac{b-1}{\sqrt{a}}\right)x^2+o(x^2)\right)}\\ & =1\end{array}$$

当 $a\ne 0$

解得 $(b-1)\sqrt{a}=0$ ，且 $\frac{1}{2}\left(\sqrt{a}+\frac{b-1}{\sqrt{a}}\right)=1$

故 $b=1$，$a=4$

当 $a=0$，极限不成立

3. 用定义法证明 $y=x^2$ 在 $(-1,2)$ 上一致连续，在 $(0,+\infty )$ 上不一致连续。

任取 $x_1,x_2\in (-1,2)$，要使不等式
$$\left|x_2^2-x_1^2\right|=\left|x_2-x_1\right|\left|x_2+x_1\right|\le 4\left|x_2-x_1\right|<\epsilon$$
成立，解得 $\left|x_2-x_1\right|<\frac{\epsilon }{4}$，取 ${\delta }_1=\frac{\epsilon }{4}$

则 $\forall x_1,x_2\in (-1,2)$，当 $\left|x_2-x_1\right|<{\delta }_1$，有 $\left|x_2^2-x_1^2\right|<\epsilon$，故在 $(-1,2)$ 上一致连续

取 $x_n=\sqrt{n}$，$y_n=\sqrt{n+1}$
lim ⁡ n → ∞ ( y n − x n ) = lim ⁡ n → ∞ ( n + 1 − n ) = lim ⁡ n → ∞ 1 n + 1 + n = 0 \begin{aligned} \mathop{\lim}\limits_{n \to \infty} (y_n - x_n) &= \mathop{\lim}\limits_{n \to \infty} \left( \sqrt{n+1} - \sqrt{n} \right) \\ &= \mathop{\lim}\limits_{n \to \infty} \frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &= 0 \end{aligned} n→∞lim​(y