求下列各极限：
- \lim\limits_{(x,y)\to(0,0)}\frac{xy}{\sqrt{2-e^{xy}}-1}.
解: 方法一: 令 z=xy, z \rightarrow 0,
(应用等价无穷小代换 (1+x)^\alpha-1 \sim \alpha x, (x \to 0))
则 \sqrt{2-e^{z}}-1 = \sqrt{1+(1-e^z)}-1 \sim \frac{1}{2}(1-e^z), 故
\begin{aligned}
\lim\limits_{(x,y)\to(0,0)}\frac{xy}{\sqrt{2-e^{xy}}-1} &= \lim\limits_{z\to0}\frac{z}{\sqrt{2-e^{z}}-1} = \lim\limits_{z\to0}\frac{z}{\frac{1}{2}(1-e^z)} \\
&= \lim\limits_{z\to0}\frac{2z}{1-e^z} = \lim\limits_{z\to0}\frac{2}{-e^z} = -2.
\end{aligned}
方法二:
利用 e^{xy}-1\sim xy((x,y)\to(0,0)),
\lim\limits_{(x,y)\to(0,0)}\frac{xy}{\sqrt{2-e^{xy}}-1}=\lim\limits_{(x,y)\to(0,0)}\frac{xy}{1-e^{xy}}\cdot(\sqrt{2-e^{xy}}+1)=-1\cdot2=-2.
-
\lim\limits_{x\to+\infty,y\to+\infty}(\frac{xy}{x^2+y^2})^x.
解：由 x^{2}+y^{2}\geq2xy,
0\le \left( \frac{xy}{x^2+y^2} \right)^x \le \left( \frac{xy}{2xy} \right)^x = \left( \frac{1}{2} \right)^x \to 0, (x \to +\infty, y \to +\infty).
故 \lim\limits_{x\to+\infty,y\to+\infty}(\frac{xy}{x^2+y^2})^x = 0. -
\lim\limits_{x\to0 ,y\to 0}\frac{x^2\sin y}{x^2+\left|y^3\right|}.
解：由 \left|y^3\right|\geq0, 可得 \frac{x^2}{x^2+\left|y^3\right|}\le1, 又 \lim\limits_{y \to 0} \sin y = 0, 所以
\lim\limits_{x\to0 ,y\to 0}\frac{x^2\sin y}{x^2+\left|y^3\right|} =0. -
\lim\limits_{(x,y)\to(1,0)}\frac{\ln(x+\mathrm{e}^y)}{\sqrt{x^2+y^2}}.
解： \lim\limits_{(x,y)\to(1,0)}\frac{\ln(x+\mathrm{e}^y)}{\sqrt{x^2+y^2}}=\frac{\ln(1+\mathrm{e}^0)}{\sqrt{1^2+0^2}}=\ln 2. -
\lim\limits_{(x,y)\to(2,0)}\frac{\tan(xy)}y.
解： (x,y)\to(2,0) 时，$xy\to0$，所以 \lim\limits_{(x,y)\to(2,0)}\frac{\tan(xy)}{xy}=1.
\lim\limits_{(x,y)\to(2,0)}\frac{\tan(xy)}y=\lim\limits_{(x,y)\to(2,0)}\frac{\tan(xy)}{xy}x=\lim\limits_{(x,y)\to(2,0)}x=2. -
\lim\limits_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)e^{x^2y^2}}.
解：
\begin{gathered} \lim\limits_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)e^{x^2y^2}} =\lim\limits_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}\cdot\frac{x^2+y^2}{e^{x^2y^2}} \\ =\frac12\cdot0=0. \end{gathered}
(1-\cos(x^2+y^2)\sim\frac12(x^2+y^2)^2, ((x,y)\to(0,0)))