下册 7.1 多元函数的极限与连续 第1题

\section*{解题过程：} （1）先求其对数的极限： $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} x^{2} y^{2} \ln \left(x^{2}+y^{2}\right)$ ． 由于 又 $$ \begin{gathered} \left|x^{2} y^{2} \ln \left(x^{2}+y^{2}\right)\right| \leqslant\left|\frac{1}{2}\left(x^{2}+y^{2}\right) \cdot \frac{1}{2}\left(x^{2}+y^{2}\right) \ln \left(x^{2}+y^{2}\right)\right|, \\ \lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}\left|\frac{1}{2}\left(x^{2}+y^{2}\right) \cdot \frac{1}{2}\left(x^{2}+y^{2}\right) \ln \left(x^{2}+y^{2}\right)\right|=0 . \end{gathered} $$ 所以 $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} x^{2} y^{2} \ln \left(x^{2}+y^{2}\right)=0$ ，进而 $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}\left(x^{2}+y^{2}\right)^{x^{2} y^{2}}=\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} e^{x^{2} y^{2} \ln \left(x^{2}+y^{2}\right)}=\mathrm{e}^{0}=1$ ． （2）令 $x=r \cos t, y=r \sin t$ ，则 $(x, y) \rightarrow(0,0) \Leftrightarrow r \rightarrow 0$ ，于是 $$ \lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}\left(x^{2}+y^{2}\right)^{x^{2}+y^{2}}=\lim _{r \rightarrow 0} r^{r}=\lim _{r \rightarrow 0} \mathrm{e}^{r \ln r}=\mathrm{e}^{0}=1 $$ （3）令 $x=r \cos \theta, y=r \sin \theta$ ，则 $(x, y) \rightarrow(0,0) \Leftrightarrow r \rightarrow 0$ ，于是 $$ \lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}(x+y) \ln \left(x^{2}+y^{2}\right)=\lim _{r \rightarrow 0} r \ln r^{2}(\sin \theta+\cos \theta)=0 $$ （4）令 $x=r \cos \theta, y=r \sin \theta$ ，则 $(x, y) \rightarrow(0,0) \Leftrightarrow r \rightarrow 0$ ，于是 $$ \lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}\left(|x|^{\alpha}+|y|^{\alpha}\right) \ln \left(x^{2}+y^{2}\right)=\lim _{r \rightarrow 0} r^{\alpha} \ln r^{2}\left(\left|\sin ^{\alpha} \theta\right|+\left|\cos ^{\alpha} \theta\right|\right)=0 $$ （5）由于 $$ \left|\sin \left(x^{3}+y^{3}\right)\right| \leqslant\left|x^{3}+y^{3}\right|=\left|x+y\left\|x^{2}+y^{2}-x y|\leqslant 2| x+y\right\| x^{2}+y^{2}\right| . $$ 于是 $\displaystyle \left|\frac{\sin \left(x^{3}+y^{3}\right)}{x^{2}+y^{2}}\right| \leqslant 2|x+y|$ ，而 $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} 2|x+y|=0$ ，所以 $\displaystyle \lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\sin \left(x^{3}+y^{3}\right)}{x^{2}+y^{2}}=0$ ． （6）由于 $$ 0 \leqslant \frac{\sqrt{|x-y|}}{x^{2}+y^{2}}\left|\sin \left(x^{2}+y^{2}\right)\right| \leqslant \frac{\sqrt{|x-y|}}{x^{2}+y^{2}}=\frac{\sqrt{x}}{\left|x^{2}\right|} \frac{1}{1+\left|\frac{y}{x}\right|^{2}} \sqrt{\left|1-\frac{y}{x}\right|} \text {, 且 } \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}}\left[\frac{\sqrt{x}}{\left|x^{2}\right|} \frac{1}{1+\left|\frac{y}{x}\right|^{2}} \sqrt{\left|1-\frac{y}{x}\right|}\right]=0 \text {, } $$ 所以 $$ \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}} \frac{\sqrt{|x-y|}}{x^{2}+y^{2}} \sin \left(x^{2}+y^{2}\right)=0 $$ （7）由于 $$ 0 \leqslant \frac{x^{2}+y^{2}}{|x|+|y|} \leqslant \frac{x^{2}+y^{2}+2|x y|}{|x|+|y|}=|x|+|y| \text {, 且 } \lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}(|x|+|y|)=0 \text {, } $$ 所以 $$ \lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2}+y^{2}}{|x|+|y|}=0 $$ （8） $\displaystyle \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}}\left(\cos \frac{y}{x}\right)^{\frac{x^{3}}{x+y^{3}}}=\lim _{\substack{x \rightarrow a \\ y \rightarrow a}} \mathrm{e}^{\frac{x^{3}}{x+y^{3}} \ln \left(1+\cos \frac{y}{x}-1\right)}=\lim _{\substack{x \rightarrow a \\ y \rightarrow a}} \mathrm{e}^{\frac{x^{3}}{x+y^{3}}\left(\cos \frac{y}{x}-1\right)}$ ． 又 $$ \begin{aligned} \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}}\left(\cos \frac{y}{x}-1\right) \frac{x^{3}}{x+y^{3}} & =-2 \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}}\left[\frac{x^{3}}{x+y^{3}} \sin ^{2}\left(\frac{y}{2 x}\right)\right]=-2 \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}}\left(\frac{x^{3}}{x+y^{3}} \cdot \frac{y^{2}}{4 x^{2}}\right) \\ & =-\frac{1}{2} \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}} \frac{x y^{2}}{x+y^{3}}=-\frac{1}{2} a^{2} \end{aligned} $$ 故 $$ \lim _{\substack{x \rightarrow \infty \\ y \rightarrow a}}\left(\cos \frac{y}{x}\right)^{\frac{x^{3}}{x+y^{3}}}=\mathrm{e}^{-\frac{1}{2} a^{2}} $$ （9）令 $\displaystyle u=\sqrt{x+\frac{1}{y}}$ ，则 $\lim _{\substack{x \rightarrow+x \\ y \rightarrow 0^{+}}} u=+\infty$ ．因为 $$ 0 \leqslant\left(x^{2}+\frac{1}{y^{2}}\right) \mathrm{e}^{-\sqrt{x+\frac{1}{y}}} \leqslant 2\left(\sqrt{x+\frac{1}{y}}\right)^{4} \mathrm{e}^{-\sqrt{x+\frac{1}{y}}}=2 u^{4} \mathrm{e}^{-u}=\frac{2 u^{4}}{\mathrm{e}^{u}} $$ 又 $\displaystyle \lim _{u \rightarrow+\infty} \frac{2 u^{4}}{\mathrm{e}^{u}}=0$ ，所以 $\displaystyle \lim _{\substack{x \rightarrow+\infty \\ y \rightarrow 0^{+}}}\left(x^{2}+\frac{1}{y^{2}}\right) \mathrm{e}^{-\sqrt{x+\frac{1}{y}}}=0$ ． （10）由于 $\displaystyle \lim _{\substack{x \rightarrow+\infty \\ y \rightarrow a}} \frac{x^{2}}{x+y} \ln \left(1+\frac{1}{x y}\right)=\lim _{\substack{x \rightarrow+\infty \\ y \rightarrow a}} \frac{x^{2}}{x+y} \frac{1}{x y}=\frac{1}{a}$ ，所以 $\displaystyle \lim _{\substack{x \rightarrow+\infty \\ y \rightarrow a}}\left(1+\frac{1}{x y}\right)^{\frac{x^{2}}{x+y}}=\mathrm{e}^{\frac{1}{a}}$ ． （11）由于 $\displaystyle \left|\left(\frac{x y}{x^{2}+y^{2}}\right)^{x^{2}}\right| \leqslant \frac{1}{2^{x^{2}}}$ ，且 $\displaystyle \lim _{x \rightarrow \infty} \frac{1}{2^{x^{2}}}=0$ ，所以 $\displaystyle \lim _{\substack{x \rightarrow+\infty \\ y \rightarrow+\infty}}\left(\frac{x y}{x^{2}+y^{2}}\right)^{x^{2}}=0$ ．