Identify limit behaviour for $f(x)=\frac{1}{\cosh x}+\log \left(\frac{\cosh x}{1+\cosh x}\right)$ for $x \to \pm \infty$. Show that $f(x) \ge 0$ I'm not sure what it means to identify the limit behaviour? If it means "just" to find the limit, then I do not have any idea. The function is very complicated.
Also, I can't use derivatives to show that $f(x) \ge 0$. So I'm not sure what to do. I think only rewriting of the function expression is allowed.
 A: Using the inequality $\log t\geq \dfrac{t-1}{t}$ for $t>0$, we obtain
$$\log\left(\frac{\cosh x}{1+\cosh x}\right)\geq\frac{\frac{-1}{1+\cosh x}}{\frac{\cosh x}{1+\cosh x}}=-\frac{1}{1+\cosh x} $$
Hence,
$$f(x)=\frac{1}{1+\cosh x}+\log\left(\frac{\cosh x}{1+\cosh x}\right)\geq 0$$
As for the limits,
$$\cosh x=\frac{e^x+e^{-x}}{2}\to\infty\,\,(x\to\pm\infty)$$
So
$$\frac{1}{\cosh x}\to 0\,\,(x\to\pm\infty)\\
\log\left(\frac{\cosh x}{1+\cosh x}\right)=\log\left(\frac{1}{\frac{1}{\cosh x}+1}\right)\to\log(1)=0\,\,(x\to\pm\infty)
 $$
and $\lim_{x\to\pm\infty}f(x)=0$. We can also observe that $f$ is an even function which means its graph is symmetric about the $y$-axis.
A: $$f(x)=\frac{1}{\cosh (x)}+\log \left(\frac{\cosh (x)}{1+\cosh (x)}\right)=\frac{1}{\cosh (x)}-\log \left(1+\frac{1}{\cosh (x)}\right)$$ Since $x$ is large, $\cosh(x)$ is large, so let $\cosh(x)=\frac 1 \epsilon$ to face
$$\epsilon-\log(1+\epsilon)=\frac{\epsilon ^2}{2}-\frac{\epsilon ^3}{3}+O\left(\epsilon ^4\right)$$ making
$$f(x)\sim \frac{1}{2}\frac{1}{\cosh^2 (x)}\sim 2e^{-2x}\to 0$$
A: The limit behavior does not just mean the limits. Rather, it means the asymptotic function form of  $f(x)=\frac{1}{\cosh x}+\log \left(\frac{\cosh x}{1+\cosh x}\right)$ in the limits $x \to \pm \infty$. 
Note that $\text{sech} (x) \to e^{-|x|}$ as $x\to \pm\infty$ and we have 
$$f(x) = \text{sech} (x )- \ln(1+\text{sech}(x)) $$
$$=e^{-|x|} - (e^{-|x|}-\frac12e^{-2|x|}+O(\frac12e^{-3|x|}))\to \frac12e^{-2|x|}$$
Thus, its limit behavior is of $f(x)\to \frac12e^{-2|x|}$ as $x\to\pm\infty$.
To show $f(x) \ge 0$, we start with the inequality $e^y = 1+y +\frac12 y^2 + … \ge 1+y$ for $y\ge0$. Then, take log of both sides to get
$$y\ge \ln(1+y)$$
Substitute $y = \frac1{\cosh(x)}$ into above inequality to obtain
$$f(x) = \frac{1}{\cosh x}-\log \left(1+\frac1{\cosh x}\right)\ge 0$$
A: We have $\cosh x \to \infty$ for $x\to \pm\infty$, thus$^1$:
$$f(x)
=\frac1{\cosh x}+\ln \left(\frac{\cosh x}{1+\cosh x}\right)
=z+\ln \frac{1}{1+z}
= g(z)
$$
with $z=1/\cosh x$ and we have to study $g(z)$ for $z\to 0^+$:
$$\begin{align}
z+\ln \frac{1}{1+z} &= z - \ln(1+z)\\
&\stackrel{(1)}\geqslant  z-((1+z)-1) = 0\\
\end{align}$$
(1) follows from $\ln x \leqslant x-1$ i.e. $-\ln x \geqslant -(x-1)$.

$^1$Assuming you denoted the Natural logarithm as $\log$.
