Using the definition of a limit, prove that 2n/3n+1=2/3 where n is any positive number?

Question

it's a limits problem.
lim 2n / (3n+1) =2/3
n->oo

it's 2n/(3n+1) =2/3 sry.

how would you prove using inductive proof methods?

lim ->oo 2n/(3n+1)=2/3 does this make more sense? if so, how will you prove this using the inductive proofs method?

Answer 1

Let f(n) = 2n/(3n + 1)

lim 2n/(3n+1) =2/3
n -> infinity

And you want to solve this using the definition of a limit? Here goes:
Recall that the definition of a limit to infinity converging goes as follows:

For every ε > 0 there exists S > 0 such that for all real numbers n > S, we have |f(n) - L| < ε

Choose S = (1 - 3E)/(9E). Then, if

N > S,
n > (1 - 3E)/(3E)
3n > (1 - 3E)/3E
3n > 1/3E - 3E/3E
3n > 1/(3E) - 1
3n + 1 > 1/(3E), therefore

|3n + 1| > 1/(3E). Multiply both sides by 3,

|3(3n+1)| > 1/E. Take the reciprocal of both sides (flips the inequality),

|1 / 3(3n + 1)| < E
| 2n/(3n+1) - (2/3) | < E

Hencing proving the limit.

Answer 2

Given a positive epsilon e you need to gaurantee that there is a N so that if n>N then |2n/(3n+1) -2/3|
Choose N large enough to be bigger than 2/(9e). Then n>2/(9e) so 9n>2/e so 9n+3>2/e so e(9n+3)>2 so e>2/(9n+3) so e>[6n+2-6n]/ (9n+3) so e> (6n+2)/(9n+3) - 6n/(9n+3) so {simplify fractions here!} e> 2/3 - 2n/(3n+1) so e> |2n/(3n+1) - 2/3| QED

Where did all that come from? I worked the problem backwards, from the goal |2n/(3n+1) - 2/3|

Answer 3

Hmmm... calculus homework..

The definition of a_n -> L is:
For all epsilon, there exists an N such that for n>N
|a_n-L|
You can think of a definition like that as a recipe for proving the statement.. Any quantity with a "for all" (like epsilon) is handed to you. Any statement with a "there exists", it's your job to find.

A proof consists of working out what N will do in terms of epsilon (which is given to you) .

A typical proof will be written out in the following way:

Let epsilon>0 be given. Let N be *@*@*@* some expression involving epsilon

Let n>N. Now we calculate

|a_n - L| < ** lots of algebra **
< ** lots more algebra **
< epsilon


The hard bit of all of this is working out what *@*@*@* is.

To do this, start by calculating |a_n - L|. Put it in as simple a form as possible. Then ask yourself the question. If someone has handed me an epsilon, how large would n need to be to guarantee that the quantity you just worked out is less than epsilon.

In this case |a_n - L| = | (2n)/(3n+1) - 2/3 |

Putting this over a common denominator, this is

|6n - (6n+2)| / (9n+3) = 2/(9n+3).

So now the question is: how large does n have to be to guarantee that 2/(9n+3)< epsilon.

You can figure this out with a bit of algebra....

Good luck...

Answer 4

lim 2n / (3n+1) = 2/3
n->oo

The value of the function APPROACHES 2/3 as n approaches ∞. The LIMIT = 2/3 , the function never does, no matter how large n is, and certainly does not equal 2/3 for small n, so the proof you desire is not possible.

Answer 5

| (4 - 2n)/(3n +2) + 2/3 | < ε => | (4 - 2n)/(3n +2) + 2/3 | 3 |3n+2|< 3 |3n+2| ε , multiply both sides by 3 |3n+2| => 16 < |9εn +6ε| , simply both sides for n -> infiinity, |9εn +6ε| > 0 so n > (16 - 6ε) / 9ε So choose N = (16 - 6ε) / 9ε and n > N = (16 - 6ε) / 9ε , and go backwards => ε |3| |3n+2| > |16| = | -6n + 12 + 6n + 4 | => therefore, ε > | (4 - 2n)/(3n +2) + 2/3 | ,divide both sides by |3| |3n+2| and simplify => therefore, lim{n to infinity} (4 - 2n)/(3n +2) = -2/3

Answer 6

Think of the problem like this:

If you were to input an extremely large number for n, what would happen? For example let's use n=10,000.
2(10,000)/(3(10,000)+1)
=20,000/30,001
=2/3 (approximately)
What would happen if you input 1,000,000 for n? You would get the same approximation.
In your case you are approaching infinity from the left, which really doesn't make a difference in this case. The best way I learned how to do limits was to use extreme situations like the one above.

Answer 7

choose any eps > 0 and then prove that for all n > some N , the 2/3 - expression is always smaller than eps.