# Limits/Absolute_Value/Inequalities Proof

1. Aug 31, 2011

### Heute

Hi folks. We're working towards understanding limits in my math class and, as part of that, here's one of our exercises:

1. The problem statement, all variables and given/known data

Find a natural number N so that absValue( (2n+3)/(4n+5)-1/2 )<.01 for all n > N

2. Relevant equations

3. The attempt at a solution

After some fiddling around with the absolute value, I have very good reasons to suspect that absValue( (2x+3)/(4x+5)-1/2 )<.01 for all real numbers x greater than or equal to 11.25. So, that would make N 11 assuming n is also a natural number.

By substituting 12 (the next natural number greater than N) for n in the original expression absValue( (2n+3)/(4n+5)-1/2 ), it simplifies to ~.0094 which is obviously less than .01.

It is fairly easy to show that the terms of a sequence given by (2n+3)/(4n+5) are strictly decreasing, so it makes sense then that if ( (2n+3)/(4n+5)-1/2 )<.01 then the same will be true for n+1 and n+2 and so on.

We also know that the terms of the sequence given by (2n+3)/(4n+5) are always positive.

So, at this point I know these three things:

1. When n=12, absValue( (2n+3)/(4n+5)-1/2 )<.01
2. The sequence where the nth term is given by (2n+3)/(4n+5) is strictly decreasing
3. The sequence where the nth term is given by (2n+3)/(4n+5) is always positive

But I don't think this is enough to prove that absValue( (2n+3)/(4n+5)-1/2 )<.01 for all n > 12

I've been thinking about using induction to prove it, but I'm not sure how to go about doing so.

2. Aug 31, 2011

### LCKurtz

You could prove that if an > 0, an+1 ≤ an, and aN ≤ c for some positive integer N, then an ≤ c for all N ≥ n by induction. But that's getting pretty pedantic. If you are in a first year Calculus class learning limits I would guess you could just state it as obvious.