上册 1.2 函数极限 第5题

\section*{解题过程：} （1） $\displaystyle \lim _{n \rightarrow \infty} \sin \left(\pi \sqrt{n^{2}+\alpha}\right)=\lim _{n \rightarrow \infty}(-1)^{n} \sin \left(\pi \sqrt{n^{2}+\alpha}-n \pi\right)=\lim _{n \rightarrow \infty}(-1)^{n} \sin \frac{\alpha \pi}{\sqrt{n^{2}+\alpha}+n}=0$ ． （2） $\displaystyle \lim _{n \rightarrow \infty} \sin ^{2}\left(\pi \sqrt{n^{2}+n}\right)=\lim _{n \rightarrow \infty} \sin ^{2}\left(\pi \sqrt{n^{2}+n}-\pi n\right)=\lim _{n \rightarrow \infty} \sin ^{2}\left(\frac{\pi}{\sqrt{1+n^{-1}}+1}\right)=\lim _{n \rightarrow \infty} \sin ^{2} \frac{\pi}{2}=1$ ． （3） $\displaystyle \lim _{n \rightarrow \infty}(\sqrt{n+\sqrt{n}}-\sqrt{n})=\lim _{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n}}=\frac{1}{2}$ ． （4） $\lim _{n \rightarrow \infty}(\sqrt{n+2}-2 \sqrt{n+1}+\sqrt{n})$ $$ =\lim _{n \rightarrow \infty}[(\sqrt{n+2}-\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n})]=\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n+2}+\sqrt{n+1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}\right)=0 $$ （5） $\lim _{x \rightarrow+\infty} \sqrt{x^{3}}(2 \sqrt{x}-\sqrt{x+1}-\sqrt{x-1})$ $$ =-\lim _{x \rightarrow+\infty} \sqrt{x^{3}}\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}-\frac{1}{\sqrt{x-1}+\sqrt{x}}\right)=-\lim _{x \rightarrow+\infty} \sqrt{x^{3}} \frac{\sqrt{x-1}-\sqrt{x+1}}{(\sqrt{x+1}+\sqrt{x})(\sqrt{x-1}+\sqrt{x})} $$ $\displaystyle =\lim _{x \rightarrow+\infty} \sqrt{x^{3}} \frac{2}{(\sqrt{x+1}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x})(\sqrt{x-1}+\sqrt{x})}=\frac{1}{4}$. （6） $\displaystyle \lim _{n \rightarrow \infty}(\sqrt{n+\sqrt{n+2 \sqrt{n}}}-\sqrt{n})=\lim _{n \rightarrow \infty} \frac{\sqrt{n+2 \sqrt{n}}}{\sqrt{n+\sqrt{n+2 \sqrt{n}}}+\sqrt{n}}=\lim _{n \rightarrow \infty} \frac{\sqrt{1+2 \sqrt{n^{-1}}}}{\sqrt{1+\sqrt{n^{-1}+\sqrt{n^{-3}}}}+1}=\frac{1}{2}$ ． （7） $\displaystyle \lim _{x \rightarrow+\infty}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})=\lim _{x \rightarrow+\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim _{x \rightarrow+\infty} \frac{\sqrt{1+\sqrt{x^{-1}}}}{\sqrt{1+\sqrt{x^{-1}+\sqrt{x^{-3}}}}+1}=\frac{1}{2}$ ． （8） $\displaystyle \lim _{n \rightarrow \infty} n\left(\sqrt{n^{2}+1}-\sqrt{n^{2}-1}\right)=\lim _{n \rightarrow \infty} n \frac{2}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=\lim _{n \rightarrow \infty} \frac{2}{\sqrt{1+\frac{1}{n^{2}}}+\sqrt{1-\frac{1}{n^{2}}}}=1$ ． （9） $\lim _{n \rightarrow \infty} \sqrt{n}\left(\sqrt[4]{n^{2}+1}-\sqrt{n+1}\right)$ $$ \begin{aligned} & =\lim _{n \rightarrow \infty} n\left(\sqrt[4]{1+\frac{1}{n^{2}}}-\sqrt{1+\frac{1}{n}}\right)=\lim _{n \rightarrow \infty} n\left[\left(1+o\left(\frac{1}{n}\right)\right)-\left(1+\frac{1}{2} \frac{1}{n}++o\left(\frac{1}{n}\right)\right)\right] \\ & =\lim _{n \rightarrow \infty} n\left[-\frac{1}{2} \frac{1}{n}+o\left(\frac{1}{n}\right)\right]=-\frac{1}{2} . \end{aligned} $$ （10） $\lim _{x \rightarrow \infty} x\left(\sqrt[3]{x^{3}+x}-\sqrt[3]{x^{3}-x}\right)$ $$ =\lim _{x \rightarrow \infty} x^{2}\left(\sqrt[3]{1+\frac{1}{x^{2}}}-\sqrt[3]{1-\frac{1}{x^{2}}}\right)=\lim _{x \rightarrow \infty} x^{2}\left[\left(1+\frac{1}{3} \frac{1}{x^{2}}\right)-\left(1-\frac{1}{3} \frac{1}{x^{2}}\right)+o\left(\frac{1}{x^{2}}\right)\right]=\frac{2}{3} . $$ （11） $\displaystyle \lim _{x \rightarrow \infty}\left(\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}\right)=\lim _{x \rightarrow \infty} x\left(\sqrt[6]{1+\frac{1}{x}}-\sqrt[6]{1-\frac{1}{x}}\right)=\lim _{x \rightarrow x} x\left[\left(1+\frac{1}{6} \frac{1}{x}\right)-\left(1-\frac{1}{6} \frac{1}{x}\right)+o\left(\frac{1}{x}\right)\right]=\frac{1}{3}$ ． （12） $\displaystyle \lim _{x \rightarrow \infty}\left(\sqrt[5]{x^{5}+x^{4}}-\sqrt[5]{x^{5}-x^{4}}\right)=\lim _{x \rightarrow \infty} x\left(\sqrt[5]{1+\frac{1}{x}}-\sqrt[5]{1-\frac{1}{x}}\right)=\lim _{x \rightarrow \infty} x\left[\left(1+\frac{1}{5} \frac{1}{x}\right)-\left(1-\frac{1}{5} \frac{1}{x}\right)+o\left(\frac{1}{x}\right)\right]=\frac{2}{5}$ ． （13）方法 1 ：令 $\displaystyle t=\frac{1}{x}$ ，则 $$ \begin{aligned} & \lim _{x \rightarrow+\infty}\left[\sqrt[n]{\left(a_{1}+x\right)\left(a_{2}+x\right) \cdots\left(a_{n}+x\right)}-x\right] \\ & =\lim _{x \rightarrow+\infty} x\left[\sqrt[n]{\left(\frac{a_{1}}{x}+1\right)\left(\frac{a_{2}}{x}+1\right) \cdots\left(\frac{a_{n}}{x}+1\right)}-1\right]=\lim _{t \rightarrow 0^{+}} \frac{\sqrt[n]{\left(a_{1} t+1\right)\left(a_{2} t+1\right) \cdots\left(a_{n} t+1\right)}-1}{t} \\ & =\left.\left[\sqrt[n]{\left(a_{1} t+1\right)\left(a_{2} t+1\right) \cdots\left(a_{n} t+1\right)}\right]^{\prime}\right|_{t=0}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n} . \end{aligned} $$ 方法 2：由于 $\displaystyle \mathrm{e}^{\frac{1}{n} \sum_{i=1}^{n} \ln \left(\frac{a_{i}}{x}+1\right)}-1 \sim \frac{1}{n} \sum_{i=1}^{n} \ln \left(\frac{a_{i}}{x}+1\right)$ ，所以 $$ \lim _{x \rightarrow+\infty}\left[\sqrt[n]{\left(a_{1}+x\right)\left(a_{2}+x\right) \cdots\left(a_{n}+x\right)}-x\right] $$ $$ \begin{aligned} & =\lim _{x \rightarrow+\infty} x\left[\sqrt[n]{\left(\frac{a_{1}}{x}+1\right)\left(\frac{a_{2}}{x}+1\right) \cdots\left(\frac{a_{n}}{x}+1\right)}-1\right]=\lim _{x \rightarrow+\infty} x\left[\mathrm{e}^{\frac{1}{n} \sum_{i=1}^{n} \ln \left(\frac{a_{i}}{x}+1\right)}-1\right] \\ & =\lim _{x \rightarrow+\infty} x \frac{1}{n} \sum_{i=1}^{n} \ln \left(\frac{a_{i}}{x}+1\right)=\lim _{x \rightarrow+\infty} \frac{1}{n} \sum_{i=1}^{n} a_{i} \ln \left(\frac{a_{i}}{x}+1\right)^{\frac{x}{a_{i}}}=\frac{1}{n} \sum_{i=1}^{n} a_{i} \end{aligned} $$ （14） $\lim _{n \rightarrow \infty}\left[\left(a_{1} \sqrt{n+1}-a_{2} \sqrt{n+2}\right)+\left(a_{3} \sqrt{n+3}-a_{4} \sqrt{n+4}\right)+\cdots+\left(a_{2 k-1} \sqrt{n+2 k-1}-a_{2 k} \sqrt{n+2 k}\right)\right]$ $$ \begin{aligned} = & \lim _{n \rightarrow \infty}\left[\left(a_{1} \sqrt{n+1}-a_{1} \sqrt{n+2}\right)+\left(a_{1} \sqrt{n+2}-a_{2} \sqrt{n+2}\right)+\left(a_{3} \sqrt{n+3}-a_{3} \sqrt{n+4}\right)+\right. \\ & \left.\left(a_{3} \sqrt{n+4}-a_{4} \sqrt{n+4}\right)+\cdots+\left(a_{2 k-1} \sqrt{n+2 k-1}-a_{2 k-1} \sqrt{n+2 k}\right)+\left(a_{2 k-1} \sqrt{n+2 k}-a_{2 k} \sqrt{n+2 k}\right)\right] \\ = & \lim _{n \rightarrow \infty}\left[-a_{1} \frac{1}{\sqrt{n+1}+\sqrt{n+2}}+\left(a_{1}-a_{2}\right) \sqrt{n+2}-a_{3} \frac{1}{\sqrt{n+3}+\sqrt{n+4}}+\left(a_{3}-a_{4}\right) \sqrt{n+4}+\cdots+\right. \end{aligned} $$ $$ \begin{aligned} & \left.\left(-a_{2 k-1} \frac{1}{\sqrt{n+2 k-1}+\sqrt{n+2 k}}\right)+\left(a_{2 k-1}-a_{2 k}\right) \sqrt{n+2 k}\right] \\ = & \lim _{n \rightarrow \infty}\left[\left(a_{1}-a_{2}\right) \sqrt{n+2}+\left(a_{3}-a_{4}\right) \sqrt{n+4}+\cdots+\left(a_{2 k-1}-a_{2 k}\right) \sqrt{n+2 k}\right] \\ = & \lim _{n \rightarrow \infty}\left\{\left[-\left(a_{3}-a_{4}\right)-\cdots-\left(a_{2 k-1}-a_{2 k}\right)\right] \sqrt{n+2}+\left(a_{3}-a_{4}\right) \sqrt{n+4}+\cdots+\left(a_{2 k-1}-a_{2 k}\right) \sqrt{n+2 k}\right\} \\ = & \lim _{n \rightarrow \infty}\left\{\left(a_{3}-a_{4}\right)[\sqrt{n+4}-\sqrt{n+2}]+\cdots+\left(a_{2 k-1}-a_{2 k}\right)[\sqrt{n+2 k}-\sqrt{n+2}]\right\} \\ = & \lim _{n \rightarrow \infty}\left[\left(a_{3}-a_{4}\right) \frac{2}{\sqrt{n+4}+\sqrt{n+2}}+\cdots+\left(a_{2 k-1}-a_{2 k}\right) \frac{2(k-1)}{\sqrt{n+2 k}+\sqrt{n+2}}\right]=0 . \end{aligned} $$