Question

利用洛必达法则求函数极限

求大神解答一下2题谢谢

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Answer 1

(2)
∫(0->1) xe^(-2x) dx

=-(1/2)∫(0->1) x de^(-2x)
=-(1/2)[ xe^(-2x)]|(0->1) + (1/2)∫(0->1) e^(-2x) dx
=-(1/2)e^(-2) -(1/4)[e^(-2x)]|(0->1)
=-(1/2)e^(-2) -(1/4)e^(-2) + 1/4
=1/4 - (3/4)e^(-2)
//
y(0) =4, y'(0) =2
y''+12y'+36y=0
The aux.equation
r^2+12r+36=0
(r+6)^2=0
r=-6
let
y= (A+Bx)e^(-6x)

y(0)=4
=>A=4
y= (A+Bx)e^(-6x)
=(4+Bx)e^(-6x)
y'=[-6(4+Bx) +B]e^(-6x)
y'(0)=2
-24+B=2
B=26
ie
y=(4+26x)e^(-6x)

Answer 2

lim<x→1>lnx/(x-1) (0/0)
= lim<x→1>(1/x)/1 = 1