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anonymous
 5 years ago
prove lim an =0 iff lim an=0. i have the if part, not the only if part. if lim an = 0 then lim an=0 and an <= an<= an.
but how do you prove if lim an = 0 then lim an = 0
anonymous
 5 years ago
prove lim an =0 iff lim an=0. i have the if part, not the only if part. if lim an = 0 then lim an=0 and an <= an<= an. but how do you prove if lim an = 0 then lim an = 0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using the sandwich theorem there. but the converse i cant figure out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is an in this context? a function?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like an = (1)^n / n^2 for n = 1,2,3...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So basically, prove: $$\lim a_n = 0 \iff \lim a_n = 0$$ Correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How do you define a limit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well the <= direction i can prove using sandwich theorem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for each epsilon there exists an N such that if n>N then an  L  < e

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, here we have an  0 < e, and we need to use that to prove an  0 < e. These are trivially equivalent since an  0 = an = an.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean we have to prove  an  0  < e

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so there are two cases

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0proof : Suppose lim an = 0 . This implies that for any e>0 there exists an N such that if n > N we have  an  0  < e

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we will show that for e* > 0 there exists an N* such that if n>N* then  an  0  < e*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, suppose, to the contrary, that $$\lim a_n = 0 \implies \lim a_n = 0$$ is false. $$\textrm{There must therefore exist an }\varepsilon\textrm{ and }N\textrm{ such that:}$$ $$a_N  L >= \varepsilon$$ But L is zero in our case, so this is equivalent to writing: $$a_N > \varepsilon$$ Which in turn is equivalent to: $$a_N > \varepsilon$$. But this contradicts our thesis that $$\lim a_n = 0.$$ Therefore, by contradiction, the original statement is true.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let e* = e from before, and let N* = N from before in our assumption. if n> N* we have  an  0 =  an  0 < e

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cant you do it directly , taking the case when an>0 and an < 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea, I think this is can be done with a simple direct proof as you said.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nomo, your proof looks good

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0davis, theres a big wrinkle in the direct proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, what if an is both positive and negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well someone tried to prove earlier this way if an>= 0 then lim an = lim an =0 , and if an< 0 then lim an = lim an = lim an = 0 ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cantorset, my proof has a small mistake. It should read: $$\textrm{``For }every\;\;\varepsilon\textrm{ there is an }N\textrm{...''}$$

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no what you wrote is correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there exists an epsilon such that for all n, if n > N then  aN  L  > epsilon, the negation of the limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the proof looks like a direct consequence of the definition of the limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0proof : Suppose lim an = 0 . This implies that for any e>0 there exists an N such that if n > N we have  an  0  < e . but an  0  =  an  0  so we can substitute that into the hypothesis, and we get our conclusion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but your proof works as well
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