Epslion Proof for limit of a sequence

[GNelson]

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 $$\epsilon$$>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.

 

[rootX]

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?

 

[GNelson]

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

 

[lurflurf]

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

 

[lurflurf]

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

 

[Rainbow Child]

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

 

[Rainbow Child]

Oupps! lurflurf was faster!

 

[GNelson]

Thanks guys working it through now.

 

[GNelson]

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.

 

[GNelson]

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

 

[Rainbow Child]

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 $N=max\left\{0,-\frac{1}{2}\,\ln(\epsilon/2)\right\}$