Question

高数求解大一高数？

Answer 1

Answer 2

lim(x->0) (cos2x)^(3/x^2)
=lim(x->0) e^[(3/x^2)*ln(cos2x)]
=e^lim(x->0) [3ln(cos2x)]/(x^2)
=e^lim(x->0) [3ln(1+cos2x-1)]/(x^2)
=e^lim(x->0) [3(cos2x-1)]/(x^2)
=e^lim(x->0) [-(3/2)*x^2]/(x^2)
=e^(-3/2)

Answer 3

(1)
x->0
e^x =1+x+o(x)
e^x +x=1+2x+o(x)
lim(x->0) ( e^x +x)^(1/x)
=lim(x->0) ( 1+2x)^(1/x)
=e^2
(5)
x->0
cos2x =1 -(1/2)(2x)^2 +o(x^2) = 1- 2x^2 +o(x^2)
lim(x->0) (cos2x)^(3/x^2)
=lim(x->0) (1- 2x^2)^(3/x^2)
=e^(-6)

追问

ox

是什么

你可以手写吗

我真的看不到

哭了

Answer 4

解：由题意可求得过点(x,y)的切线方程是Y-y=y'(X-x),即Y=y'X+y-xy'
∴此切线与纵坐标轴交点是(0,y-xy')
==>切点(x,y)到纵坐标轴间的切线段长是√[x²+(xy')²]
∵一曲线上点(x,y)的切线自切点到纵坐标轴间的切线段有定长2
∴√[x²+(xy')²]=2
==>x²+(xy')²=4
==>x²+x²y'²=4
==>x²(y'²+1)=4
故曲线应满足的微分方程是 x²(y'²+1)=4