(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 im stumped, our textbook is very poor so are the notes.

Any help is welcome thanks in advanced.

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# Homework Help: Epslion Proof for limit of a sequence

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