2023数学分析【南昌大学】

计算

1. 求极限 $\underset{n\to \infty }{\lim }\left(\frac{1}{\sqrt{n^2+1^2}}+\frac{1}{\sqrt{n^2+2^2}}+\cdots +\frac{1}{\sqrt{n^2+n^2}}\right)$。
2. 求极限 $\underset{x\to 0}{\lim }\frac{\cos x-e^{-\frac{x^2}{2}}}{x^2{\tan }^{2}x}$。
3. 设 $f(x)={\int }_1^x{e}^{-t^2}\mathrm{d}t$，求 ${\int }_0^1{x}^{2}f(x)\mathrm{d}x$。
4. 设 $u=x^2+y^2+z^2$，$z=f(x,y)$ 且 $x^2+y^2+z^2=3xyz$，考虑使用隐函数，求 $u_{xx}$。

解答 1：

$$\begin{array}{rl}\underset{n\to \infty }{\lim }\left(\frac{1}{\sqrt{n^2+1^2}}+\frac{1}{\sqrt{n^2+2^2}}+\cdots +\frac{1}{\sqrt{n^2+n^2}}\right)& =\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^n\frac{1}{\sqrt{1+{\left(\frac{k}{n}\right)}^2}}\\ & ={\int }_0^1\frac{1}{\sqrt{1+x^2}}\mathrm{d}x\\ & ={\int }_0^{\frac{\pi }{4}}\frac{\sec t\left(\sec t+\tan t\right)}{\sec t+\tan t}\mathrm{d}t\\ & =\ln \left|\sec t+\tan t\right|{|}_0^{\frac{\pi }{4}}\\ & =\ln \left(\sqrt{2}+1\right)\end{array}$$

解答 2：

lim ⁡ x → 0 cos ⁡ x − e − x 2 2 x 2 tan ⁡ 2 x = lim ⁡ x → 0 cos ⁡ x − e − x 2 2 x 4 = lim ⁡ x → 0 ( 1 − x 2 2 ! + x 4 4 ! + o ( x 4 ) ) − ( 1 − x 2 2 + ( − x 2 2 ) 2 2 ! + o ( x 4 ) ) x 4 = lim ⁡ x → 0 1 6 x 4 + o ( x 4 ) x 4 = 1 6 \begin{align*} \mathop{\lim }\limits_{x \to 0} \frac{\cos x - e^{ - \frac{x^2}{2}}}{x^2 \tan^2 x} &= \mathop{\lim }\limits_{x \to 0} \frac{\cos x - e^{ - \frac{x^2}{2}}}{x^4} \\ &= \mathop{\lim }\limits_{x \to 0} \frac{\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + o(x^4)\right) - \left(1 - \frac{x^2}{2} + \frac{\left(- \frac{x^2}{2}\right)^2}{2!} + o(x^4)\right)}{x^4} \\ &= \mathop{\lim }\limits_{x \to 0} \frac{\frac{1}{6} x^4 + o(x^4)}{x^4} \\ &= \frac{1}{6} \\ \end{align*} x→0lim​x2tan2x