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.

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.
