couple questions about limit of sequences

[rocadelpunk]

hi.

just got a couple questions to clarify some limit stuff haven't done in a while.

first up is sequence An= lim as n goes to infin of cos(nPi)/n

i figured since cos(np) alternates b/t -1,1 that the sequence would diverge, but book says answer is 0...so for alternating stuff cases would i just use squeeze theorem say something like since the lim|An| = 0 due to -1/n<|cos(npi)|<1/n thus lim An = 0

and then for this one where An = (-pi)^n/5^n, just say same thing..abs value of lim = normal limit so it'd be small over big = 0



also would seq An = (1/4)^n _ 3^n/2 div as lim goes to infin? ..(1/4)^n goes to 0 and then the n/2 would go to infin so just div..


An = n^100/e^n would = 0 since e^n will grow quicker


and how would you show that An= (1 +2/n)^n/2 is = e

is e = to lim as n goes to infin (1+1/n)^n so i assume you just say the 2 doesn't matter due to it going to infin.

 

[Capone]

It's been awhile since I had calc 2 but without busting out the pencil and paper, I'm guessing you have the first problem wrote wrong because cos (nPi)/n would always be the cos Pi because the n's would cancel. I'm guessing it's supposed to be (cos nPi)/n. and this would not always alternate between 1 and -1 because it would be cos 1pi/1 which would be -1, then (cos 2pi)/2 which is 1/2, -1/3 etc so It's going to keep getting smaller on each side, so thats where the 0 comes from. Give me a min to look at the others.

 

[Capone]

second one looks alright because its just going to be alternating pi over infinity. so anything over infinty is 0.

I'm not sure what the problem is in the third one. What is the _ in there?

e^n does not grow faster than n^100. 2^100 is way bigger than e^2

 

[Pepsi90919]

what

 

[chuckywang]

Originally posted by: rocadelpunk

and how would you show that An= (1 +2/n)^n/2 is = e

is e = to lim as n goes to infin (1+1/n)^n so i assume you just say the 2 doesn't matter due to it going to infin.

no, the 2 does matter.

Let k = n/2, so the sequence becomes (taking n->infinity is the same as taking k->infinity)

(1+1/k)^k, which you know as k->infinity is e.

 

[rocadelpunk]

thanks for the replies.