x→0,y→0[2-√(xy+4)]/xy的极限.x→2,y→0sin(xy)/y的极限
第一题:分子分母同乘以[2+√(xy+4)],则分母变为xy[2+√(xy+4)],分子变为4-xy-4=-xy,分子与分母的xy消去,即可求出极限为:-1/4第二题:分子用等价无穷小代换为xy,与分母约去y,可得极限为:2
举一反三
- 下列方程中( )是一阶线性微分方程。 A: \( 2{x^2}yy' = {y^2} + 1 \) B: \( xy' + {y \over x} - x = 0 \) C: \( \cos y + x\sin y { { dy} \over {dx}} = 0 \) D: \( y'' + xy' = 4{x^2} + 1 \)
- 分解因式()x()3()y()-()2()x()2()y()2()+()xy()3()正确的是A.()xy()(()x()+()y())()2()B.()xy()(()x()2()﹣()2()xy()+()y()2())()C.()xy()(()x()2()+2()xy()﹣()y()2())()D.()xy()(()x()﹣()y())()2
- 下列方程中( )是微分方程。 A: \( x{y^3} + 2{y^2} + {x^2}y = 0 \) B: \( {y^2} + xy - y = 0 \) C: \( x + {y^2} = 0 \) D: \( dy + ydx = 0 \)
- (多选)以下平面弹性体的位移或形变状态不可能存在的是 A: 位移分量$u = {k_1}\left( {{x^2} + {y^2}} \right),v = {k_2}xy$(${k_1},{k_2}$为常数) B: ${\varepsilon _x} = k\left( {{x^2} + {y^2}} \right),{\varepsilon _y} = k{y^2},{\gamma _{xy}} = 2kxy$(${k \ne 0}$) C: ${\varepsilon _x} = 0,{\varepsilon _y} = 0,{\gamma _{xy}} = kxy$(${k \ne 0}$) D: ${\varepsilon _x} = ax{y^2},{\varepsilon _y} = b{x^2}y,{\gamma _{xy}} = cxy$($a \ne 0,b \ne 0,c \ne 0$)
- 下列微分方程中,( )是齐次方程。 A: \( xy' = y(\ln y - \ln x) \) B: \( xy' + {y \over x} - x = 0 \) C: \( y' + {y \over x} = {1 \over { { x^2}}} \) D: \( y - y' = 1 + xy' \)
内容
- 0
设E(X) =E(Y)= 1/3 , E(XY)= 0, D(X) =D(Y)=2/9 , , 则ρXY =____
- 1
\({\lim_{x\to 0}}\)\({\lim_{y\to 0}}\)\(\frac{xy}{x^2+y^2}\) <br/>______
- 2
已知|x|=3,|y|=2,xy<0,则x+y=________.
- 3
设 $D=\{(x,y)|x^2+y^2\le 4, y\ge 0\}$,则二重积分 $\iint_D xy^2dxdy=$______ .
- 4
下列微分方程是线性微分方程的是()。 A: x(y’)<sup>2</sup>+y=e<sup>x</sup> B: xy"+xy’+y=cosx C: y<sup>3</sup>y"+y’+2y=0 D: y"+2y"+y<sup>2</sup>=0