Limit of function of hyperbolic How can I - without using derivatives - find the limit of the function
$f(x)=\frac{1}{\cosh(x)}+\log \left(\frac{\cosh(x)}{1+\cosh(x)} \right)$
as $x \to \infty$ and as $x \to -\infty$?
We know that $\cosh(x) \to \infty$ as $x \to \pm \infty$ thus $\frac{1}{\cosh(x)} \to 0$ as $x \to \pm \infty$.
And I imagine that $\frac{\cosh(x)}{1+\cosh(x)} \to 1$ as $x \to \pm \infty$ thus $\log\left(\frac{\cosh(x)}{1+\cosh(x)}\right) \to 0$ as $x \to \pm \infty$.
Is this approach sufficiently formal?
Any help is appreciated.

$$
\begin{aligned}
\lim _{x\to \infty }\left(\frac{1}{\cosh \left(x\right)}+\ln\left(\frac{\cosh \left(x\right)}{1+\cosh \left(x\right)}\right)\right)
& = \lim _{x\to \infty }\left(\frac{1+\ln \left(\frac{\left(\frac{e^x+e^{-x}}{2}\right)}{\left(\frac{e^x+e^{-x}}{2}\right)+1}\right)\left(\frac{e^x+e^{-x}}{2}\right)}{\left(\frac{e^x+e^{-x}}{2}\right)}\right)
\\& = \lim _{x\to \infty }\left(\frac{2+\ln \:\left(\frac{e^x+e^{-x}}{e^x+e^{-x}+2}\right)\left(e^x+e^{-x}\right)}{e^x+e^{-x}}\right)
\\& \approx \lim _{x\to \infty }\left(\frac{2+\ln \:\left(\frac{e^x}{e^x+2}\right)\left(e^x\right)}{e^x}\right)
\\& \approx \lim _{x\to \infty }\left(\frac{2+0\cdot \left(e^x\right)}{e^x}\right)
\\& = \color{red}{0}
\end{aligned}
$$
