# Epslion Proof for limit of a sequence

• GNelson
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.
GNelson

## Homework Statement

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

## The Attempt at a Solution

My attempt at the solution is as follows.

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

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

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

we have 1-(e^2n-1/(e^2n+1)) < $$\epsilon$$

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

Any help is welcome thanks in advanced.

GNelson said:

## Homework Statement

we have 1-(e^2n-1/(e^2n+1)) < $$\epsilon$$

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

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

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

you we so close
(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

Try this

$$\left|\frac{e^{2n}-1}{e^{2n}+1}-1\right|=\left|\frac{-2}{e^{2n}+1}\right|=\frac{2}{e^{2n}+1}<\frac{2}{e^{2n}}=2\,e^{-2n}<\epsilon$$

and choose
$$N=max\left\{0,-\frac{1}{2}\,\ln\frac{\epsilon}{2}\right\}$$

Oupps! lurflurf was faster!

Thanks guys working it through now.

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.

And I got the problem thank you, However I am still curious about the derivation of N

There is no general method to choose $N$. One starts with the definition $|a_n-l|<\epsilon$ and tries to find $N$. In the problem at hand, since we don't know if $-\frac{1}{2}\,\ln(\epsilon/2)$ is positive we have to write $N=max\left\{0,-\frac{1}{2}\,\ln(\epsilon/2)\right\}$

## 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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