Limits/Absolute_Value/Inequalities Proof

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In summary, the problem is to find a natural number N such that absValue( (2n+3)/(4n+5)-1/2 )<.01 for all n > N. After some experimentation and using the fact that the sequence (2n+3)/(4n+5) is strictly decreasing and always positive, it is found that N=11 satisfies the equation. However, this is not enough to prove it for all n > 11. Induction may be used to prove it, but it may not be necessary for a first year Calculus class learning limits.
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Hi folks. We're working towards understanding limits in my math class and, as part of that, here's one of our exercises:

Homework Statement



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

Homework Equations



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.
 
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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.
 

1. What is a limit in mathematics?

A limit in mathematics is the value that a function approaches as the input of the function gets closer and closer to a specific value. It is denoted by the symbol "lim" and is used to describe the behavior of a function near a certain point.

2. How do you prove a limit using the epsilon-delta definition?

To prove a limit using the epsilon-delta definition, you must show that for any positive value of epsilon, there exists a corresponding positive value of delta such that the distance between the input and the specific value is less than delta, the distance between the output and the limit is less than epsilon. In other words, you must show that the function can be made as close as desired to the limit value by choosing a small enough interval around the specific value.

3. What is an absolute value?

An absolute value is the distance of a number from zero on the number line. It is always a positive value and is denoted by vertical bars around the number. For example, the absolute value of -5 is 5, and the absolute value of 3 is also 3.

4. How do you prove an absolute value inequality?

To prove an absolute value inequality, you must show that the statement holds true for both the positive and negative versions of the given inequality. This can be done by breaking the inequality into two separate inequalities and solving each one individually. For example, to prove |x| < 5, you must show that both x < 5 and x > -5 are true.

5. What is the difference between an open and closed interval in inequalities?

In inequalities, an open interval does not include the endpoints, while a closed interval does include the endpoints. For example, in the inequality x > 1, the interval would be written as (1, ∞) for an open interval and [1, ∞) for a closed interval. This distinction is important when solving absolute value inequalities as it affects the solutions.

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