✔ 最佳答案
Q1~Q3都是級數嗎?
2010-04-16 23:06:45 補充：
Q1:
lim(n->∞) a(n)=lim(n->) (2n+1)/[1- 3sqrt(n)]
= lim(n->∞) [2sqrt(n)+ 1/sqrt(n)]/[1/sqrt(n) - 3]
= -∞ not equal to 0
so, Σ a(n) div.
Q2:
lim(n->∞) a(n)= lim(n->∞) (1+ 7/n)^n= e^7
Σ a(n) div.
Q3:
lim(n->∞) ln(1+ 1/n)^n= ln(e)= 1
Σ a(n) div.
Q4:
∫(-∞~∞) 2x/(x^2+1)^2 dx
= -1/(x^2+1) sub. for x=-∞~∞
=0-0=0
Note: lim(x->+/-∞) 1/(x^2+1)=0
Q5:
∫[0~1] xLn(x) dx
= (x^2/2)Lnx |[0~1] -∫[0~1] x/2 dx
= (x^2/2)Ln(x) - x^2/4 sub. x=0~1
= -1/2
Note: lim(x->0+) x^2 Ln(x)=0
2010-04-16 23:55:12 補充：
sqrt(n)=√n
sub. 為substitute(代值)的縮寫.
2010-04-17 12:56:56 補充：
Q4:∫(∞,-∞)2xdx/(x^2+1)^2=lim(t->-∞)∫(t~0) 2x/(x^2+1)^2 dx+lim(u->∞)∫[0~u)2x/(x^2+1)^2 dx
=lim(t->-∞) [-1+1/(t^2+1)] +lim(u->∞) [-1/(u^2+1)+1]=-1+1=0
Q5:∫[0~1] xLn(x)dx=lim(t->0+)∫[t~1] xLn(x)dx= lim(t->0+) (x^2/4)(2Lnx-1)|[t~1]
=lim(t->0+)[ -1/4+(t^2/4)(2Lnt-1)]= -1/4 (原作答 -1/2錯)
2010-04-17 12:59:28 補充：
lim(t->0+) [t^2*Lnt]=lim(u->∞) (-Lnu)/(u^2)=(by L' rule)lim(u->∞) -1/(2u^2) = 0
