tobytoby1995

[Limit] Question


Please check in Q2 and 3?
The limit doesn't exist??? Strange Odd!!?

Please help on Q4-6!!!!!

[ 本帖最後由 tobytoby1995 於 2011-4-5 04:37 PM 編輯 ]

MrDon

I don't think it's strange. At many time, the limit doesn't exist.

Simon

2. Limit should exist.
3. Limit should exist.

hinokari

Try to multiply or divide n to let one constant there and others is in form of k/n

tobytoby1995

yes but can't do that in Q2/3? No limit? denominator =0 or infinity??/

夏娜醬大好

4. The limit doesn't exist

5,6 題是不是n→∞？
如果n→∞：

5. e-1/(1!) -2/(2!) -3/(3!) = e - 5/3
6. Let S be the answer.
   -(3^4)S = -3^5/(5!) +3^6/(6!) -3^7/(7!) + ...
   -(3^4)S = (-3)^5/(5!) +(-3)^6/(6!) +(-3)^7/(7!) + ...
   -(3^4)S = e^(-3) - 1- (-3)^1/(1!) - (-3)^2/(2!) - (-3)^3/(3!) - (-3)^4/(4!)
   -(3^4)S = e^(-3) - 11/8
             S = [-e^(-3) + 11/8] / 81

都是用e^x = Sum[n=0 to ∞] (x^n)/(n!) 這條公式

Simon

sorry, i've just ignored the denominator
yes, should be "limit does not exist as the whole thing does not converge"

夏娜醬大好

引用:

原帖由 Simon 於 2011-4-5 04:47 PM 發表
2. Limit should exist.
3. Limit should exist.

何出此言？
2,3題都是「The limit doesn't exist.」

(無事了,樓上已更正)

[ 本帖最後由 夏娜醬大好 於 2011-4-5 05:35 PM 編輯 ]

cc2001ym

7)  e   唔洗解 by definition
8)  lim n->oo  ( 1+ 1/n) ^n  * ( 1+1/n)^ 10
     = e *  ( 1+ 0) = e
9)   lim n -> oo  ( (n-4)/(n+1))^ (n-3)
    = lim n -> oo ( 1 +  -5 / (n+1)  ) ^ n+1  * ( 1 - 5/ (n+1) )^ (-4)
    = e ^ (-5)  *(1+0)^ (-4)
    = e^ (-5)
唔知 岩唔岩 請指正

夏娜醬大好

引用:

原帖由 cc2001ym 於 2011-4-5 07:33 PM 發表
7)  e   唔洗解 by definition
8)  lim n->oo  ( 1+ 1/n) ^n  * ( 1+1/n)^ 10     
  = e *  ( 1+ 0)
  = e
9)  lim n -> oo  ( (n-4)/(n+1))^ (n-3)   
  = lim n -> oo ( 1 +  -5 / (n+1)  ) ^ n+1  * ( 1 - 5/ (n ...

沒有錯
但樓主似乎沒有叫我們做第7-9題喔

tobytoby1995

Nevermind, Thanks you guys for your generous and immense assistance ! :)