## 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

1. anonymous

using the sandwich theorem there. but the converse i cant figure out

2. anonymous

What is an in this context? a function?

3. anonymous

a sequence

4. anonymous

like an = (-1)^n / n^2 for n = 1,2,3...

5. anonymous

So basically, prove: $$\lim a_n = 0 \iff \lim |a_n| = 0$$ Correct?

6. anonymous

How do you define a limit?

7. anonymous

well the <= direction i can prove using sandwich theorem

8. anonymous

for each epsilon there exists an N such that if n>N then |an - L | < e

9. anonymous

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

you mean we have to prove | |an| - 0 | < e

11. anonymous

so there are two cases

12. anonymous

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

now we will show that for e* > 0 there exists an N* such that if n>N* then | |an| - 0 | < e*

14. anonymous

ok so far?

15. anonymous

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

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

cant you do it directly , taking the case when an>0 and an < 0

18. anonymous

Yea, I think this is can be done with a simple direct proof as you said.

19. anonymous

20. anonymous

davis, theres a big wrinkle in the direct proof

21. anonymous

well, what if an is both positive and negative

22. anonymous

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

cantorset, my proof has a small mistake. It should read: $$\textrm{For }every\;\;\varepsilon\textrm{ there is an }N\textrm{...''}$$

24. anonymous

no what you wrote is correct

25. anonymous

there exists an epsilon such that for all n, if n > N then | aN - L | > epsilon, the negation of the limit

26. anonymous

the proof looks like a direct consequence of the definition of the limit

27. anonymous

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

but your proof works as well