$\frac{1}{\cosh{x}}+\log\left ( \frac{\cosh{x}}{1+\cosh{x}} \right )$ for $x \rightarrow \pm \infty$ has a limit Show from the definition of a limit that $$\frac{1}{\cosh{x}}+\log\left ( \frac{\cosh{x}}{1+\cosh{x}} \right )$$ for $x \rightarrow \pm \infty$ has a limit.
My attempt
This one is really tough for me. I know the limit is 0 and I know that $f(x) \rightarrow b$ for $x \rightarrow \infty$ if $\forall \epsilon>0 \exists M >0: \left \| f(x)-b \right \|<\epsilon$ for all $x \in A, x>M$ ($A$ is the domain of the map $f$ and $A \subset \mathbb{R}^m$). From this I should be able to prove the following but have no idea on how to do it and where to apply my inequalities. I am pretty sure I have to use the fact that $\frac{x}{1-x}\leq \log(1-x)\leq x$ and $0<\frac{1}{\cosh{x}}\geq 2e^{-x}$. When I get the result for $x\rightarrow \infty$ I should get it for $x\rightarrow -\infty$ because of symmetri around the $y$-axis.
 A: First solution: Re-write your expression as
$$
\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right).
$$
Then
$$
\left| {\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right)} \right| \le \frac{1}{{\cosh x}} + \log \left( {1 + \frac{1}{{\cosh x}}} \right) \le \frac{2}{{\cosh x}}
$$
for all real $x$. Let $\varepsilon>0$ be arbitrary. Then if $\left| x \right| > \log \left( {\frac{4}{\varepsilon }} \right)$,
$$
\frac{2}{{\cosh x}} \le 4e^{ - |x|}  < \varepsilon .
$$
Second solution: A more optimal solution may be given as follows. Note that
$$
\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right) = \frac{1}{{\cosh ^2 x}}\int_0^1 {\frac{t}{{1 + \frac{t}{{\cosh x}}}}dt} .
$$
Then
$$
\left| {\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right)} \right| \le \frac{1}{{\cosh ^2 x}}\int_0^1 {tdt}  = \frac{1}{2}\frac{1}{{\cosh ^2 x}}.
$$
Let $\varepsilon>0$ be arbitrary. Then if $\left| x \right| > \frac{1}{2}\log \left( {\frac{2}{\varepsilon }} \right)$,
$$
\frac{1}{2}\frac{1}{{\cosh ^2 x}} \le 2e^{ - 2\left| x \right|}  < \varepsilon .
$$
