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：

$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\cos x-e^{-\frac{x^2}{2}}}{x^2{\tan }^{2}x}& =\underset{x\to 0}{\lim }\frac{\cos x-e^{-\frac{x^2}{2}}}{x^4}\\ & =\underset{x\to 0}{\lim }\frac{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4)\right)-\left(1-\frac{x^2}{2}+\frac{{\left(-\frac{x^2}{2}\right)}^2}{2!}+o(x^4)\right)}{x^4}\\ & =\underset{x\to 0}{\lim }\frac{\frac{1}{6}{x}^4+o(x^4)}{x^4}\\ & =\frac{1}{6}\end{array}$$

解答 3：

$$\begin{array}{rl}{\int }_0^1\mathrm{d}x{\int }_1^x{x}^2{e}^{-t^2}\mathrm{d}t& =-{\int }_0^1\mathrm{d}t{\int }_0^t{x}^2{e}^{-t^2}\mathrm{d}x\\ & =-{\int }_0^1\frac{t^3}{3}{e}^{-t^2}\mathrm{d}t\\ & =-\frac{1}{6}{\int }_0^1{t}^2{e}^{-t^2}\mathrm{d}(t^2)\\ & =\frac{1}{6}\left(t^2{e}^{-t^2}+e^{-t^2}\right){|}_0^1\\ & =\frac{1}{3}{e}^{-1}-\frac{1}{6}\end{array}$$

解答 4：

$$u_x=2x+2z{z}_x=3yz+3xy{z}_x$$

得到：

$$z_x=\frac{2x-3yz}{3xy-2z}$$

$$\begin{array}{rl}{z}_{xx}& =\frac{2+2z_x^2-6y{z}_x}{3xy-2z}\\ & =\frac{1}{{\left(3xy-2z\right)}^3}[2\left(9x^2{y}^2-12xyz+4z^2\right)+2(4x^2-12xyz+9y^{2}z\\ & \end{array}$$