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.

You just need to know that
$$
\lim_{x\to\infty}\cosh x=\infty
$$
so also
$$
\lim_{x\to\infty}\frac{1}{\cosh x}=0
$$
Once you have this, half of your assignment is done; now you can do
$$
\lim_{x\to\infty}\log\frac{\cosh x}{1+\cosh x}=
\lim_{x\to\infty}\log\frac{1}{\dfrac{1}{\cosh x}+1}=\log\frac{1}{0+1}=0
$$
Since $\cosh(-x)=\cosh x$, the limit at $-\infty$ is the same.
Why is $\lim_{x\to\infty}\cosh x=\infty$?
Consider
$$
\cosh x=\frac{e^x+e^{-x}}{2}=\frac{e^x}{2}+\frac{1}{2e^x}
$$
and the fact that $\lim_{x\to\infty}e^x=\infty$
