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.

For $x\to \pm\infty$, 
$$\lim\limits_{x\to \pm\infty}\underbrace{\log\bigg(\frac{\cos h x}{1+\cos h x}\bigg)}_{\to 0}+\lim\limits_{x\to \pm\infty}\underbrace{\sec h x}_{\to 0}$$
