多元函数的极限、连续与微分1

$1.求极限：（1）\lim_{(x,y) \to (0,0)}(1+x^2+y^2)^{\frac{xy}{x^2+y^2}}（2）\lim_{x \to \infty , y \to \infty}\frac{x+y}{x^2-xy+y^2}$

• \(（1）\)

\[\lim_{(x,y) \to (0,0)}(1+x^2+y^2)^{\frac{xy}{x^2+y^2}}=\lim_{(x,y) \to (0,0)}e^{\frac{xy}{x^2+y^2}ln(1+x^2+y^2)}=e^{\lim_{(x,y) \to (0,0)}\frac{xy}{x^2+y^2}ln(1+x^2+y^2)} \]

\[e^{\lim_{(x,y) \to (0,0)}\frac{xy}{x^2+y^2}ln(1+x^2+y^2)}=e^{\lim_{(x,y) \to (0,0)}xy\cdot\lim_{(x,y) \to (0,0)}\frac{ln(1+x^2+y^2)}{x^2+y^2}}=e^{0 \cdot 1}=1 \]

• \(（2）\)

\[x^2+y^2 \geq 2|xy| \]

\[|\frac{x+y}{x^2-xy+y^2}| \leq |\frac{x+y}{2xy-xy}|\leq\frac{1}{|y|}+\frac{1}{|x|} \]

\[\lim_{x \to \infty , y \to \infty}\frac{1}{|y|}+\frac{1}{|x|}=0 \]

• \(由夹逼准则，原极限等于0\)

\(2.求极限：\lim_{x \to +\infty,y \to +\infty}(x^2+y^2)e^{-(x+y)}\)

\[\lim_{x \to +\infty,y \to +\infty}(x^2+y^2)e^{-(x+y)}=\lim_{x \to +\infty,y \to +\infty}\frac{x^2}{e^x}e^{-y}+\frac{y^2}{e^y}e^{-x}=0+0=0 \]

\(3.设函数\)

\[f(x,y)=\left \{ \begin{array}{rcl}&\frac{x^2y}{x^4+y^2} , (x,y) \neq 0 \\ &0,(x,y) = 0 \end{array} \right. \]

\(证明在(0,0)处偏导数存在，但在该点函数不连续。\)

\[f_x(0,0)=\lim_{x \to 0}\frac{f(x,0)-f(0,0)}{x-0}=0 \]

\[f_y(0,0)=\lim_{y \to 0}\frac{f(0,y)-f(0,0)}{y-0}=0 \]

• \(所以f(x,y)在(0,0)处的偏导数均存在\)
• \(对于直线y=x:\)

\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^4+y^2}=\lim_{x \to 0}\frac{x^3}{x^4+x^2}=0 \]

• \(对于抛物线y=x^2:\)

\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^4+y^2}=\lim_{x \to 0}\frac{x^4}{2x^4}=\frac{1}{2} \]

• \(因此可找到两条不同的路径使得(x,y) \to (0,0)时的极限不同，故极限\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^4+y^2}不存在，从而函数在(0,0)不连续\)

\(4.设函数\)

\[f(x,y)=\left \{ \begin{array}{rcl}&\frac{x^2y}{x^2+y^2} , (x,y) \neq 0 \\ &0,(x,y) = 0 \end{array} \right. \]

\(证明该函数在原点连续，偏导数存在，但不可微。\)

• \(连续和可偏导的证明这里省略，没什么难度\)
• \(这里只证明不可微\)

• \(假设f(x,y)在(0,0)可微，应有极限\)

\[\lim_{(x,y) \to (0,0)}\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=\lim_{(x,y) \to (0,0)}\frac{x^2y}{(x^2+y^2)^{\frac{3}{2}}}=0 \]

• \(对于直线y=x:\)

\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{(x^2+y^2)^{\frac{3}{2}}}=\lim_{x\to 0}\frac{x^3}{\sqrt{8}x^3}=\frac{1}{2\sqrt{2}} \]

• \(对于直线y=0:\)

\[\lim_{(x,y) \to (0,0)}\frac{x^2y}{(x^2+y^2)^{\frac{3}{2}}}=0 \]

• \(所以\lim_{(x,y) \to (0,0)}\frac{x^2y}{(x^2+y^2)^{\frac{3}{2}}}不存在，与假设矛盾\)
• \(故原函数在(0,0)不可微\)

\(5.设f(x,y)=|x-y| \varphi(x,y)，其中\varphi(x,y)在(0,0)处连续。证明f(x,y)在(0,0)处可微的充分必要条件是\varphi(0,0)=0。\)

• \(充分性\)

\[|\frac{|x-y|\varphi(x,y)-0}{\sqrt{x^2+y^2}}|\leq\frac{|x||\varphi(x,y)|+|y||\varphi(x,y)|}{\sqrt{x^2+y^2}} \]

\[\frac{|x||\varphi(x,y)|+|y||\varphi(x,y)|}{\sqrt{x^2+y^2}}\leq \frac{|x||\varphi(x,y)|}{\sqrt{x^2}}+\frac{|y||\varphi(x,y)|}{\sqrt{y^2}}=2|\varphi(x,y)| \to 0 \]

• \(所以原函数在(0,0)可微，充分性得证\)

• \(必要性\)
• \(假设\varphi(0,0)=a \neq 0\)
• \(对于直线y=2x:\)

\[\lim_{(x,y) \to (0,0)}\frac{|x-y|\varphi(x,y)-0}{\sqrt{x^2+y^2}}=\lim_{x \to 0}\frac{|x|\varphi(x,x)}{\sqrt{5}|x|}=\frac{a}{\sqrt{5}}\neq0 \]

• \(原函数不可微，与必要性条件矛盾，故\varphi(0,0)=0\)
• \(必要性得证\)