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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

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  1. anonymous
    • 5 years ago
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    using the sandwich theorem there. but the converse i cant figure out

  2. anonymous
    • 5 years ago
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    What is an in this context? a function?

  3. anonymous
    • 5 years ago
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    a sequence

  4. anonymous
    • 5 years ago
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    like an = (-1)^n / n^2 for n = 1,2,3...

  5. anonymous
    • 5 years ago
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    So basically, prove: $$\lim a_n = 0 \iff \lim |a_n| = 0$$ Correct?

  6. anonymous
    • 5 years ago
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    How do you define a limit?

  7. anonymous
    • 5 years ago
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    well the <= direction i can prove using sandwich theorem

  8. anonymous
    • 5 years ago
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    for each epsilon there exists an N such that if n>N then |an - L | < e

  9. anonymous
    • 5 years ago
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    Well, 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.

  10. anonymous
    • 5 years ago
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    you mean we have to prove | |an| - 0 | < e

  11. anonymous
    • 5 years ago
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    so there are two cases

  12. anonymous
    • 5 years ago
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    proof : 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

  13. anonymous
    • 5 years ago
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    now we will show that for e* > 0 there exists an N* such that if n>N* then | |an| - 0 | < e*

  14. anonymous
    • 5 years ago
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    ok so far?

  15. anonymous
    • 5 years ago
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    So, 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.

  16. anonymous
    • 5 years ago
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    let e* = e from before, and let N* = N from before in our assumption. if n> N* we have | |an| - 0| = | an - 0| < e

  17. anonymous
    • 5 years ago
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    cant you do it directly , taking the case when an>0 and an < 0

  18. anonymous
    • 5 years ago
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    Yea, I think this is can be done with a simple direct proof as you said.

  19. anonymous
    • 5 years ago
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    nomo, your proof looks good

  20. anonymous
    • 5 years ago
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    davis, theres a big wrinkle in the direct proof

  21. anonymous
    • 5 years ago
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    well, what if an is both positive and negative

  22. anonymous
    • 5 years ago
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    well 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 ,

  23. anonymous
    • 5 years ago
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    cantorset, my proof has a small mistake. It should read: $$\textrm{``For }every\;\;\varepsilon\textrm{ there is an }N\textrm{...''}$$

  24. anonymous
    • 5 years ago
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    no what you wrote is correct

  25. anonymous
    • 5 years ago
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    there exists an epsilon such that for all n, if n > N then | aN - L | > epsilon, the negation of the limit

  26. anonymous
    • 5 years ago
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    the proof looks like a direct consequence of the definition of the limit

  27. anonymous
    • 5 years ago
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    proof : 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

  28. anonymous
    • 5 years ago
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    but your proof works as well

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