Epslion Proof for limit of a sequence

In summary, using the definition of a limit of a sequence, we can prove that as n approaches infinity, tanh(n) approaches 1. This can be shown by demonstrating that for any given epsilon, there exists a natural number N such that when n is greater than N, the absolute value of (tanh(n) - 1) is less than epsilon. By manipulating the given equation and choosing N=max\left\{0,-\frac{1}{2}\,\ln(\epsilon/2)\right\}, we can show that the limit of exp(-n) as n approaches infinity is equal to 0, which in turn proves the original statement.
  • #1
GNelson
9
0

Homework Statement



Using only the definition of a limit of a sequence prove that lim n->infinity tanh(n)=1


Homework Equations





The Attempt at a Solution



My attempt at the solution is as follows.

If 1 is the limit of the sequence then for every [tex]\epsilon[/tex]>0, there exists an number such that n>N for every n, such that we have

|tanh(n)-1|<[tex]\epsilon[/tex]
apply the appropriate hyperbolic identity I re-write this as.

|e^2n-1/(e^2n+1) -1 |< [tex]\epsilon[/tex]
as tanh(n)< 1 for every sufficiently large n

we have 1-(e^2n-1/(e^2n+1)) < [tex]\epsilon[/tex]

After this I am stumped, our textbook is very poor so are the notes.

Any help is welcome thanks in advanced.
 
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  • #2
GNelson said:

Homework Statement



we have 1-(e^2n-1/(e^2n+1)) < [tex]\epsilon[/tex]

How about factoring out e^2n, and then applying limit, and if you are missing lim n-->inf sign there?
 
  • #3


If i could evaluate it as a limit I would but its asking to prove it using the precise definition of a limit of a sequence, which means that is not an option
 
  • #4
lim n->infinity tanh(n)=1
we desire to show
for any eps>0
there exist (a natural number) N(eps)
such that whenever (a natural number) n>N(eps)
|1-tanh(n)|<eps
hint
show |1-tanh(n)|<2exp(-2n)
Then the original question is equivalent to showing
lim n->infinity exp(-n)=0
 
  • #5
you we so close
you had
(1-(e^2n-1/(e^2n+1))
((e^2n+1)/(e^2n+1)-(e^2n-1)/(e^2n+1))
((e^2n+1)-(e^2n-1))/(e^2n+1)
2/(e^2n+1)
2/(e^2n+1)<2exp(-n)
thus the hint
it should be easy to finish
 
  • #6
Try this

[tex]\left|\frac{e^{2n}-1}{e^{2n}+1}-1\right|=\left|\frac{-2}{e^{2n}+1}\right|=\frac{2}{e^{2n}+1}<\frac{2}{e^{2n}}=2\,e^{-2n}<\epsilon[/tex]

and choose
[tex]N=max\left\{0,-\frac{1}{2}\,\ln\frac{\epsilon}{2}\right\}[/tex]
 
  • #7
Oupps! lurflurf was faster! :smile:
 
  • #8
Thanks guys working it through now.
 
  • #9
I am currently in calc II. I am curious where you got what we choose N= , I can finish it with the substitution you gave me. I am just curious how one derives it.
 
  • #10
And I got the problem thank you, However I am still curious about the derivation of N
 
  • #11
There is no general method to choose [itex]N[/itex]. One starts with the definition [itex]|a_n-l|<\epsilon[/itex] and tries to find [itex]N[/itex]. In the problem at hand, since we don't know if [itex]-\frac{1}{2}\,\ln(\epsilon/2)[/itex] is positive we have to write [itex]N=max\left\{0,-\frac{1}{2}\,\ln(\epsilon/2)\right\}[/itex]
 

Related to Epslion Proof for limit of a sequence

1. What is an Epsilon Proof for the limit of a sequence?

An Epsilon Proof for the limit of a sequence is a method used to prove that a sequence converges to a specific limit. It involves choosing a small value, epsilon, and showing that all terms in the sequence after a certain point are within epsilon of the limit.

2. Why is an Epsilon Proof important in mathematics?

An Epsilon Proof is important in mathematics because it provides a rigorous way to prove the convergence of a sequence. It is also a fundamental concept in analysis and helps to understand the behavior of functions and limits.

3. How do you construct an Epsilon Proof for the limit of a sequence?

To construct an Epsilon Proof, you first choose a small value, epsilon, and then find a point in the sequence where all terms after it are within epsilon of the limit. This can be done by using the definition of limit and finding a suitable value for n, the number of terms in the sequence.

4. Can you give an example of an Epsilon Proof for the limit of a sequence?

One example of an Epsilon Proof for the limit of a sequence is the proof that the sequence {1/n} converges to 0 as n approaches infinity. By choosing any value of epsilon, we can find a point in the sequence where all terms after it are within epsilon of the limit, in this case, 0.

5. Are there any limitations to using Epsilon Proofs for limits of sequences?

One limitation of Epsilon Proofs is that they only work for sequences that have a limit. If a sequence does not have a limit, an Epsilon Proof cannot be used. Additionally, constructing Epsilon Proofs can be challenging and time-consuming, especially for more complicated sequences.

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