$1.求极限：（1）\lim_{(x,y) \to (0,0)}(1+x^2+y^2)^{\frac{xy}{x^2+y^2}}（2）\lim_{x \to \infty , y \to \infty}\frac{x+y}{x^2-xy+y^2}$
\(2.求极限：\lim_{x \to +\infty,y \to +\infty}(x^2+y^2)e^{-(x+y)}\) \(3.设函数\) \(证明在(0,0)处偏导数存在，但在该点函数不连续。\) \(4.设函数\) \(证明该函数在原点连续，偏导数存在，但不可微。\) \(5.设f(x,y)=|x-y| \varphi(x,y)，其中\varphi(x,y)在(0,0)处连续。证明f(x,y)在(0,0)处可微的充分必要条件是\varphi(0,0)=0。\)
\[\lim_{(x,y) \to (0,0)}(1+x^2+y^2)^{\frac{xy}{x^2+y^2}}=\lim_{(x,y) \to (0,0)}e^{\frac{xy}{x^2+y^2}ln(1+x^2+y^2)}=e^{\lim_{(x,y) \to (0,0)}\frac{xy}{x^2+y^2}ln(1+x^2+y^2)}
\]\[e^{\lim_{(x,y) \to (0,0)}\frac{xy}{x^2+y^2}ln(1+x^2+y^2)}=e^{\lim_{(x,y) \to (0,0)}xy\cdot\lim_{(x,y) \to (0,0)}\frac{ln(1+x^2+y^2)}{x^2+y^2}}=e^{0 \cdot 1}=1
\]
\[x^2+y^2 \geq 2|xy|
\]
\[|\frac{x+y}{x^2-xy+y^2}| \leq |\frac{x+y}{2xy-xy}|\leq\frac{1}{|y|}+\frac{1}{|x|}
\]\[\lim_{x \to \infty , y \to \infty}\frac{1}{|y|}+\frac{1}{|x|}=0
\]
\[\lim_{x \to +\infty,y \to +\infty}(x^2+y^2)e^{-(x+y)}=\lim_{x \to +\infty,y \to +\infty}\frac{x^2}{e^x}e^{-y}+\frac{y^2}{e^y}e^{-x}=0+0=0
\]
\[f_x(0,0)=\lim_{x \to 0}\frac{f(x,0)-f(0,0)}{x-0}=0
\]\[f_y(0,0)=\lim_{y \to 0}\frac{f(0,y)-f(0,0)}{y-0}=0
\]
\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^4+y^2}=\lim_{x \to 0}\frac{x^3}{x^4+x^2}=0
\]
\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^4+y^2}=\lim_{x \to 0}\frac{x^4}{2x^4}=\frac{1}{2}
\]
\[\lim_{(x,y) \to (0,0)}\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=\lim_{(x,y) \to (0,0)}\frac{x^2y}{(x^2+y^2)^{\frac{3}{2}}}=0
\]
\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{(x^2+y^2)^{\frac{3}{2}}}=\lim_{x\to 0}\frac{x^3}{\sqrt{8}x^3}=\frac{1}{2\sqrt{2}}
\]
\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{(x^2+y^2)^{\frac{3}{2}}}=0
\]
\[|\frac{|x-y|\varphi(x,y)-0}{\sqrt{x^2+y^2}}|\leq\frac{|x||\varphi(x,y)|+|y||\varphi(x,y)|}{\sqrt{x^2+y^2}}
\]\[\frac{|x||\varphi(x,y)|+|y||\varphi(x,y)|}{\sqrt{x^2+y^2}}\leq \frac{|x||\varphi(x,y)|}{\sqrt{x^2}}+\frac{|y||\varphi(x,y)|}{\sqrt{y^2}}=2|\varphi(x,y)| \to 0
\]
\[\lim_{(x,y) \to (0,0)}\frac{|x-y|\varphi(x,y)-0}{\sqrt{x^2+y^2}}=\lim_{x \to 0}\frac{|x|\varphi(x,x)}{\sqrt{5}|x|}=\frac{a}{\sqrt{5}}\neq0
\]