题目
4.计算题.-|||-(1)求极限:-|||-① y)→(0,0) x^2y^2 e^xy-|||-lim 1-cos(xy )-|||-0,-|||-② . lim _(xarrow y)dfrac (1-xy)({x)^2+(y)^2}-|||-bigcirc (3) lim 1-√(xy+1)-|||-y)→(0,0) xy-|||-④ lim _((x,y)arrow (0,0))(xsin dfrac (1)(y)+ysin dfrac (1)(x))-|||-(2)设 =sin (xy)+cos ((y)^2), 求 dfrac (partial z)(partial x),dfrac (partial z)(partial y)-|||-(3)设 =(e)^usin v =x+y =x-y, 求 dfrac (partial z)(partial x),dfrac (partial z)(partial y)-|||-(4)设 =(e)^x+y =sin t =(t)^2, 求 dfrac (dz)(dt).-|||-(5)设 =3(x)^2y+3x(y)^2-2xy, 求 dfrac ({a)^2z}(a{x)^2}, dfrac ({partial )^2z}(partial xpartial y) dfrac ({partial )^2z}(partial {y)^2}.-|||-(6)设函数 z=z(x,y) 由方程 ^x=2xy+3yz+zx 所确定,求 dfrac (partial z)(partial x) dfrac (partial z)(partial y).-|||-(7)求函数 (x,y)=(x)^3+3(y)^2-6xy+1 的极值.-|||-(8)求曲线 x=t+1 =3(t)^2, =(t)^3-1 在点(2,0,0)处的切线与法平面方程-|||-(9)求曲面 ^x=2z-xy+5 在点(1,4,0)处的切平面及法线方程.-|||-(10)交换下列积分次序:4.计算题.-|||-(1)求极限:-|||-① y)→(0,0) x^2y^2 e^xy-|||-lim 1-cos(xy )-|||-0,-|||-② . lim _(xarrow y)dfrac (1-xy)({x)^2+(y)^2}-|||-bigcirc (3) lim 1-√(xy+1)-|||-y)→(0,0) xy-|||-④ lim _((x,y)arrow (0,0))(xsin dfrac (1)(y)+ysin dfrac (1)(x))-|||-(2)设 =sin (xy)+cos ((y)^2), 求 dfrac (partial z)(partial x),dfrac (partial z)(partial y)-|||-(3)设 =(e)^usin v =x+y =x-y, 求 dfrac (partial z)(partial x),dfrac (partial z)(partial y)-|||-(4)设 =(e)^x+y =sin t =(t)^2, 求 dfrac (dz)(dt).-|||-(5)设 =3(x)^2y+3x(y)^2-2xy, 求 dfrac ({a)^2z}(a{x)^2}, dfrac ({partial )^2z}(partial xpartial y) dfrac ({partial )^2z}(partial {y)^2}.-|||-(6)设函数 z=z(x,y) 由方程 ^x=2xy+3yz+zx 所确定,求 dfrac (partial z)(partial x) dfrac (partial z)(partial y).-|||-(7)求函数 (x,y)=(x)^3+3(y)^2-6xy+1 的极值.-|||-(8)求曲线 x=t+1 =3(t)^2, =(t)^3-1 在点(2,0,0)处的切线与法平面方程-|||-(9)求曲面 ^x=2z-xy+5 在点(1,4,0)处的切平面及法线方程.-|||-(10)交换下列积分次序:
题目解答
答案
解析
步骤 1：计算极限 $\lim _{(x,y)\rightarrow (0,0)}\dfrac {1-\cos (xy)}{x^2y^2e^{xy}}$
使用泰勒展开式，$\cos(xy) = 1 - \frac{(xy)^2}{2} + O((xy)^4)$，代入极限表达式中，得到：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {1-\cos (xy)}{x^2y^2e^{xy}} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {1-(1 - \frac{(xy)^2}{2} + O((xy)^4))}{x^2y^2e^{xy}} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {\frac{(xy)^2}{2} + O((xy)^4)}{x^2y^2e^{xy}}$$
由于 $e^{xy}$ 在 $(0,0)$ 处的值为1，所以极限简化为：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {\frac{(xy)^2}{2} + O((xy)^4)}{x^2y^2} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {1}{2} + O((xy)^2) = \dfrac {1}{2}$$
步骤 2：计算极限 $\lim _{(x,y)\rightarrow (0,0)}\dfrac {1-\sqrt{xy+1}}{xy}$
使用泰勒展开式，$\sqrt{xy+1} = 1 + \frac{xy}{2} + O((xy)^2)$，代入极限表达式中，得到：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {1-\sqrt{xy+1}}{xy} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {1-(1 + \frac{xy}{2} + O((xy)^2))}{xy} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {-\frac{xy}{2} + O((xy)^2)}{xy}$$
简化得到：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {-\frac{xy}{2} + O((xy)^2)}{xy} = \lim _{(x,y)\rightarrow (0,0)}-\dfrac {1}{2} + O(xy) = -\dfrac {1}{2}$$
步骤 3：计算极限 $\lim _{(x,y)\rightarrow (0,0)}(x\sin \dfrac {1}{y}+y\sin \dfrac {1}{x})$
由于 $\sin \dfrac {1}{y}$ 和 $\sin \dfrac {1}{x}$ 在 $(0,0)$ 处的值是有限的，而 $x$ 和 $y$ 趋近于0，所以极限为0。
使用泰勒展开式，$\cos(xy) = 1 - \frac{(xy)^2}{2} + O((xy)^4)$，代入极限表达式中，得到：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {1-\cos (xy)}{x^2y^2e^{xy}} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {1-(1 - \frac{(xy)^2}{2} + O((xy)^4))}{x^2y^2e^{xy}} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {\frac{(xy)^2}{2} + O((xy)^4)}{x^2y^2e^{xy}}$$
由于 $e^{xy}$ 在 $(0,0)$ 处的值为1，所以极限简化为：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {\frac{(xy)^2}{2} + O((xy)^4)}{x^2y^2} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {1}{2} + O((xy)^2) = \dfrac {1}{2}$$
步骤 2：计算极限 $\lim _{(x,y)\rightarrow (0,0)}\dfrac {1-\sqrt{xy+1}}{xy}$
使用泰勒展开式，$\sqrt{xy+1} = 1 + \frac{xy}{2} + O((xy)^2)$，代入极限表达式中，得到：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {1-\sqrt{xy+1}}{xy} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {1-(1 + \frac{xy}{2} + O((xy)^2))}{xy} = \lim _{(x,y)\rightarrow (0,0)}\dfrac {-\frac{xy}{2} + O((xy)^2)}{xy}$$
简化得到：
$$\lim _{(x,y)\rightarrow (0,0)}\dfrac {-\frac{xy}{2} + O((xy)^2)}{xy} = \lim _{(x,y)\rightarrow (0,0)}-\dfrac {1}{2} + O(xy) = -\dfrac {1}{2}$$
步骤 3：计算极限 $\lim _{(x,y)\rightarrow (0,0)}(x\sin \dfrac {1}{y}+y\sin \dfrac {1}{x})$
由于 $\sin \dfrac {1}{y}$ 和 $\sin \dfrac {1}{x}$ 在 $(0,0)$ 处的值是有限的，而 $x$ 和 $y$ 趋近于0，所以极限为0。