anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
using the sandwich theorem there. but the converse i cant figure out
anonymous
  • anonymous
What is an in this context? a function?
anonymous
  • anonymous
a sequence

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
like an = (-1)^n / n^2 for n = 1,2,3...
anonymous
  • anonymous
So basically, prove: $$\lim a_n = 0 \iff \lim |a_n| = 0$$ Correct?
anonymous
  • anonymous
How do you define a limit?
anonymous
  • anonymous
well the <= direction i can prove using sandwich theorem
anonymous
  • anonymous
for each epsilon there exists an N such that if n>N then |an - L | < e
anonymous
  • 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.
anonymous
  • anonymous
you mean we have to prove | |an| - 0 | < e
anonymous
  • anonymous
so there are two cases
anonymous
  • 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
anonymous
  • anonymous
now we will show that for e* > 0 there exists an N* such that if n>N* then | |an| - 0 | < e*
anonymous
  • anonymous
ok so far?
anonymous
  • 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.
anonymous
  • 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
anonymous
  • anonymous
cant you do it directly , taking the case when an>0 and an < 0
anonymous
  • anonymous
Yea, I think this is can be done with a simple direct proof as you said.
anonymous
  • anonymous
nomo, your proof looks good
anonymous
  • anonymous
davis, theres a big wrinkle in the direct proof
anonymous
  • anonymous
well, what if an is both positive and negative
anonymous
  • 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 ,
anonymous
  • anonymous
cantorset, my proof has a small mistake. It should read: $$\textrm{``For }every\;\;\varepsilon\textrm{ there is an }N\textrm{...''}$$
anonymous
  • anonymous
no what you wrote is correct
anonymous
  • anonymous
there exists an epsilon such that for all n, if n > N then | aN - L | > epsilon, the negation of the limit
anonymous
  • anonymous
the proof looks like a direct consequence of the definition of the limit
anonymous
  • 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
anonymous
  • anonymous
but your proof works as well

Looking for something else?

Not the answer you are looking for? Search for more explanations.