Proving the sequence $\left\{\frac{e^n-e^{-n}}{e^n+e^{-n}}\right\}$ converges I am asked to show that the sequence $$\left\{\frac{e^n-e^{-n}}{e^n+e^{-n}}\right\}$$ converges using the definition of convergence. Thus, I am trying to do this by using the definition of convergence, i.e., I am looking for a value for $n$ that will make this true. However, I am having some trouble since this involves $e$. 
I get to this point,
$$\frac{1}{e}<\frac{e^n+e^{-n}}{e^n-e^{-n}}$$
but I am not sure how to continue to get $n$ by itself. Any suggestions? 
 A: Note that as @carmichael561 said, you really need to find what this tends to in order to figure out an $n$ for which 
$$
\bigg|\frac{e^n-e^{-n}}{e^n+e^{-n}} -L\bigg|<\epsilon
$$
fortunately it's not too hard to figure out $L=1$. Then we work backwards as usual 
$$
\bigg|\frac{e^n-e^{-n}}{e^n+e^{-n}} -1\bigg|<\epsilon\implies 
\bigg|\frac{-2e^{-n}}{e^n+e^{-n}}\bigg|<\epsilon\\
\implies 2e^{-n}<\epsilon (e^n+e^{-n})\implies \frac{2}{\epsilon}<\frac{e^n+e^{-n}}{e^{-n}}\\
\implies \frac{2}{\epsilon}<e^{2n}+1
$$
then we see that we can take 
$$
n>\log(\sqrt{2/\epsilon-1})
$$
which will make sense as long as we take $\epsilon<2$.
If $\epsilon>2$, any $n\in \mathbb{N}$ will do. Note the intuition behind this, the sequence is bounded between $0$ and $1$, thus, any value will be within $2$ of $1$ trivially.
A: The limit is $1$. Let's try to solve, for $m> 0$: 
$$\bigg|\frac{1-e^{-2n}}{1+e^{-2n}} -1\bigg|< 10^{-m}$$
I don't really know but if I remember correctly if $x > 0 $
$$-2x(1-x) <\frac{1-x }{ 1+x  } - 1  = \frac{-2x }{ 1+x  }  < 
-2x 
$$
So if we set $2x = e^{-2n}<10^{-m}$ we are done.
