Proof involving limit of a general case

[Heute]

Homework Statement

Let a,b,c,d be real numbers. Assume that both c and cn+d are not equal to zero for all natural numbers n.

Prove that for every positive real number \epsilon there exits some natural number N so that n > N \Rightarrow |(an+c)/(cn+d) - (a/c)| < \epsilon

In other words, prove that the limit as n approaches infinity of the sequence defined by (an+b)/(cn+d) is a/c

Homework Equations

The Attempt at a Solution

I started by trying to figure out what N was in terms of \epsilon by working backwards from where I want to be:

|(an+b)/(cn+d)-(a/c)| < \epsilon

We can simplify what's in the absolute value brackets to get:

|(bc-ad)/(c^2n+cd)| < \epsilon (that's c^2 times n - not c to the 2n power)

And from there,

|(c^2n+cd)/(bc-ad)| > 1/\epsilon
|c^2n+cd| > |bc-ad|/\epsilon
|c(cn+d)| > |bc-ad|/\epsilon
|cn+d| > |bc-ad|/|c|\epsilon
|cn|+|d| > |bc-ad|/|c|\epsilon since |cn+d|<=|cn|+|d| by the Triangle Inequality

|cn| > |bc-ad|/|c|\epsilon-|d|
|n| > |bc-ad|/c^2\epsilon - |d/c|
n > |bc-ad|/c^2\epsilon - |d/c| since n > 0

Now, to actually go about proving the initial statement, I would start with the assumption that for a given \epsilon let N be a natural number greater than

|bc-ad|/c^2\epsilon - |d/c|

and then work backwards from there to prove that this implies

|(bc-ad)/(c^2n+cd)| < \epsilon

The problem is where I used the Triangle Inequality. While |cn+d| > |bc-ad|/|c|\epsilon implies|cn|+|d| > |bc-ad|/|c|\epsilon, the converse is NOT true. I've encountered situations like this in class, but we had actual numerical values for a,b,c, and d so I could usually use some sort of property or intuition that was more obvious to me.

By the way, I'm not very familiar with using the equation editing system on this forum so I apologize for the probably tricky to read mathematical statements. If anyone has a reference as to how to display equations better, please let me know. Thanks!

 

[micromass]

What if you also restrict n by asking n\geq -d?

This forces n+d to be positive, and thus |n+d|=n+d...

 

[Heute]

I don't think I follow you there. I tried what you said and it didn't affect my reasoning. Here:

given \epsilon > 0, let N be a natural number greater than |bc-ad|/c^(2)\epsilon-|d/c|

Assume n > N (and n >= -d). Then,

n > |bc-ad|/c^(2)\epsilon-|d/c|
n + |d/c| > |bc-ad|/c^(2)\epsilon
c^(2)n + |cd| > |bc-ad|/\epsilon
1/(c^(2)n + |cd|) < \epsilon/|bc-ad|
|bc-ad|/(c^(2)n + |cd|) < \epsilon
|bc-ad|/(|c|(|c|n + |d|)) < \epsilon (factoring out a |c|)

The problem is we need to get the denominator to |c|*|cn+d|, but the Triangle inequality suggests |cn|+|d| >= |cn+d|, and therefore 1/(|cn|+|d|) <= 1/|cn+d| meaning if we "combine" the absolute values being added in the denominator, we could get a bigger value on the right hand side of the inequality and that step cannot be justified.

 

[micromass]

Oh, I didn't see a c in front of the n. That might complicate things.

So you have

\frac{|bc-ad|}{c^2n + cd} &lt; \varepsilon

And we must make the denominator into |c^2n+cd| somehow.

The thing is, if we take n large enough, then c^2n (which is positive) becomes larger than |cd| and then |c^2n+cd|=c^2n+cd. Thus

\frac{|bc-ad|}{c^2n+cd}=\frac{|bc-ad|}{|c^2n+cd|}&lt;\varepsilon

Do you see what I'm trying to do?

 

[Heute]

I'm afraid I really do not see what you're doing there. I understand that as n becomes very large, cd becomes negligible by comparison - but I'm not sure if that's the direction you're going or - if it is - that using a concept about limits in proving a limit is a "fair move."

For what it's worth, the first inequality you wrote should have cd in absolute value bars.

I've got a feeling my issue is that I need a different value for N - something that still makes the limit true but doesn't involve the Triangle Inequality in its derivation.

 

[micromass]

I'm afraid I really do not see what you're doing there. I understand that as n becomes very large, cd becomes negligible by comparison - but I'm not sure if that's the direction you're going or - if it is - that using a concept about limits in proving a limit is a "fair move."

I'm not using a limit concept. I'm using Archimedes' axiom that arbitrary large natural numbers exist. Hence, I can take n larger enough.

For what it's worth, the first inequality you wrote should have cd in absolute value bars.

I know, but those absolute value bars are avoidable. You don't need them.

 

[Heute]

I know, but those absolute value bars are avoidable. You don't need them.

I don't know why this is true.

The thing is, if we take n large enough, then c2n (which is positive) becomes larger than |cd| and then |c2n+cd|=c2n+cd

Erm, so let me try and put this in concrete terms. Suppose c = -2 and d = 3. If n were large enough eventually

|4n-6| = 4n-6

Ok - let n = 100, and we have |394|= 394

I see what you mean. But part of my process for proving the limit includes establishing a minimum value N so that n > N implies |bc-ad|/|c^(2)n+cd| < \epsilon. So then is the right idea to specify that n > [ N = |bc-ad|/|c^(2)\epsilon| ] AND c^(2)n > |cd|?

 

[micromass]

I don't know why this is true.

Well, if you start with

n > |bc-ad|/c^(2)ϵ-(d/c)

instead of

n > |bc-ad|/c^(2)ϵ-|d/c|

then you'll obtain a version without absolute value signs.

So then is the right idea to specify that n > [ N = |bc-ad|/|c^(2)\epsilon| ] AND c^(2)n > |cd|?

Yes, that is what I'm meaning.

 

[Heute]

Ok, that helps a lot. Let me see if I can pull all of this together into a proof now.

Given some \epsilon > 0, let N be any natural number greater than |bc-ad|/c^(2)\epsilon - (d/c) so that N is also greater than d/c

Now assume that n > N. Then,

n > |bc-ad|/c^(2)\epsilon-(d/c)
n + (d/c) > |bc-ad|/c^(2)\epsilon
c^(2)n + cd > |bc - ad|/\epsilon
(c^(2)n + cd) / |bc - ad| > 1/\epsilon
|bc - ad|/(c^(2)n + cd) < \epsilon

Since N > d/c, n > d/c. Therefore (c^(2)n + cd) > 0, therefore (c^(2)n + cd) = |c^(2)n + cd| giving us

|bc - ad|/|c^(2)n + cd| < \epsilon

which can be reconfigured to give us what we need.


correct me if I'm wrong, but I think that's solid. Thanks much!

 

[micromass]

I think that's ok!