Help with Limit Proof: A(n) & 1/A(n)

[migetmonkee91]

Homework Statement

The Lim as n goes to infinity of the sequence A(n) = A, A does not equal zero

Then prove Lim as n goes to infinity of 1/a(n) = 1/A

Homework Equations

How do i attack this? I am using the epsilon definition of course, but I have no idea how to manipulate this.

The Attempt at a Solution

So far i have looked at the formal definitions of the limits of both of the sequences, and seen if they could interact anyway, and so far, i cannot find much.

any help would be greatly appreciated!

I thought about letting Epsilon being fixed for a particular delta, but have no idea how that would work in the long run, and i also tried just plain manipulation.

 

[hunt_mat]

Let b_{n}=1/A_{n} and let b=1/A, now what are the conditions that b_n->b as n->infinity?

 

[migetmonkee91]

Ah, can't help but feeling like a complete idiot after that...thank you very much

Quickly though, does that really prove it, and why does it?

I know that the definition of the limit is The entire Epsilon > 0 |Xn-L| business, so is that why? It just doesn't seem like a thorough idea, so i just want to be sure.

 

[migetmonkee91]

My idea being i still have to PROVE that |bn-B| < epsilon, which i guess I'm still stuck on.

 

[hunt_mat]

Look at:
<br /> \frac{1}{A_{n}}-\frac{1}{A}=-\frac{A_{n}-A}{A_{n}A}<br />
You know that there is an N>N_0 such that |A_{n}-A|<e. Since A_n converges it is possile to find an N such that A_N>A/2. Use these statements to complete the question.

 

[migetmonkee91]

I am still having trouble on this and i feel as if i am missing something completely obvious. I got now that |a_n-A / a_nA must be less than epsilon. So i know the numerator converges to A, but I have no idea what to do with the denominator.

 

[hunt_mat]

Choose an N>N_0 such that |A_n-A|<A^2e/2 and A_n>A/2 then
<br /> \frac{|A_{n}-A|}{A_{n}A}&lt;\frac{A^{2}\varepsilon /2}{A^{2}/2}=\varepsilon<br />
Hence...

 

[migetmonkee91]

Where did the (A^2e/2) / A^2/2 come from?

 

[hunt_mat]

It's a trick, to make:
<br /> \left|\frac{1}{A_{n}}-\frac{1}{A}\right| &lt;\varepsilon<br />

 

[migetmonkee91]

Ah i see. Is there a particular reason that was chosen? Like why A^2/2, why would i need to try to find An -A > A/2

 

[migetmonkee91]

I meant An > A/2? I just am more confused then ever.

 

[hunt_mat]

You have A_nA on the denoninator, so you have to estimate the A_n, you know that it is possible to find an N such that A_n>A/2, hence there will be a N such that A_nA>A^2/2 or 1/A_nA<2/A^2. Does this make things clearer?

 

[migetmonkee91]

I'm sorry, can we just start from the beginning on this?

So you have |1/A_n - 1/A| = |-A_n-A/An(A)| due to the definition of a limit and all of that.

I understand this much. Now we understand A_n converges to A, and thus An-A will be less then the right side, so we have to prove A_n(A) also is smaller. This because A_n>A/2, there will be an N S.t. A_nA>A^2/2 sincey oure just multiplying both sides by A. Now this is where i am still confused. You took the reciprocal in your last post but it's A^2/2 In the first post you made. If it's possible for you to take it step by step in a single post that would be greatly helpful.

 

[hunt_mat]

You have to show that
<br /> \frac{|A_{n}-A|}{|A_{n}A|}&lt;\varepsilon<br />
We know that A_n converges and that we can find an N_1 in the natural numbers such that |A_n|>|A|/2. Also there is an N_2 such that |A_n-A|<e_{1}. Chose N=max{N_1,N_2}, then:
<br /> \frac{|A_{n}-A|}{|A_{n}A|}&lt;\frac{2\varepsilon_{1}}{A^{2}}=\varepsilon<br />
Choosing
<br /> \frac{2\varepsilon_{1}}{A^{2}}=\varepsilon<br />
Hence you have proved the theorem.

 

[migetmonkee91]

The only thing now is why E1 x 2 / A^2 = Epsilon?

 

[hunt_mat]

It's just to make it look nice, you could leave it as it is and you would have still proven the theorem.