Question

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

Answer 1

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

Answer 2

What is an in this context? a function?

Answer 3

a sequence

Answer 4

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

Answer 5

So basically, prove:

$$\lim a_n = 0 \iff \lim |a_n| = 0$$

Correct?

Answer 6

How do you define a limit?

Answer 7

well the <= direction i can prove using sandwich theorem

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

so there are two cases

Answer 12

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

Answer 13

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

Answer 14

ok so far?

Answer 15

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.

Answer 16

let e* = e from before, and let N* = N from before in our assumption. if n> N* we have | |an| - 0| = | an - 0| < e

Answer 17

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

Answer 18

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

Answer 19

nomo, your proof looks good

Answer 20

davis, theres a big wrinkle in the direct proof

Answer 21

well, what if an is both positive and negative

Answer 22

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 ,

Answer 23

cantorset, my proof has a small mistake. It should read:

$$\textrm{``For }every\;\;\varepsilon\textrm{ there is an }N\textrm{...''}$$

Answer 24

no what you wrote is correct

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

but your proof works as well