Prove some laws of sequencing limits

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The discussion revolves around homework problems related to proving laws of sequencing limits. The first problem involves demonstrating that the limit of the product of two sequences equals the product of their limits, which can be shown using the epsilon-delta definition of limits. The second problem requires proving that the sequence defined by an = (-1)^n diverges, as it fails to meet the convergence criteria for any proposed limit. Participants provide insights on how to approach these proofs, emphasizing the importance of understanding the definitions and properties of limits. Overall, the conversation highlights the challenges students face in grasping these concepts and the collaborative effort to find solutions.
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Hi, I got some problems with the homework that I was assigned last week. I really got some problems to even start doing it...it's too difficult that I don't even know how to start it...the professor asked us to proof some laws...of sequencing limits when I don't even understand anything...I've discussed the problem with lots of classmates but non of them know how to do them either...please help me PROOF these problems><" thanxz in advance!
PROOF:
#1 Limit Law: lim (anbn) = (lim an ) (lim bn)
n->infinity n->infinity n->infinity


#2 an = (-1)^n (1,-1,1,-1,1,-1...) when n=0,1,2,3... Proof that it diverges
THANKS A LOT IF YOU COULD HELP ME!@@"
 
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I think this is an exercise in 'epsilontics'.
Meaning, you use the standard definition of a limes:

a = lim(n->[oo])an

if and oly if

for each epsilon > 0
there exists n0, so that
|an - a| < epsilon
for each n > n0.
 
yeah, I saw my professor writing something like thaht on the board, but I have no idea how and what that is@@"...could you help me doing the problem further?? please?
 
I've really never learned that...please help me
 
You just saw it? And didn't take notes?

OK, as for #1:
You want to show that |anbn - ab|
goes to zero, OK?
Here's a trick: Let's insert - abn + abn (which is zero):
|anbn - ab|
= |anbn - abn + abn- ab|
= |(an - a)bn + (bn - b)a|

since (an - a) and (bn - b) go to zero, the whole expression goes to zero. It's a matter of writing it down properly.
 
oh...ok, thanxz a lot arcnets...actually I did take notes...but the problem is that the professor just went through the whole thing whithout saying much about it...anyways...thank you!
uhmmm...do you have any clue on how to proof number two by chance? Anybody!?@@"
 
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the first problem looks something like this:
img11.gif

does anyone know how to proof it@@? please?! I desperately want to know it...thanxz
 
PLEASEEEEEEEEEEEEE!
 
1. arcnets already showed you how to do this, why do you keep asking about it?

2. A sequence converges to a number L if for every &epsilon; > 0 there exists an N such that n > N implies |xn-L|<&epsilon;. Given xk=(-1)k, for any L there exists infinitely many m such that |xm-L| >= 1, so the definition of convergence can't be satisfied satisfied. This prove is a little more slick if you can use the concept of a cauchy sequence which is not a difficult concept.
 
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thank you grady!
 
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