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.
 A: If you know everything then you have nothing to proove ;)
If not you could use the Euler formula
$$\cosh x := \frac12 \left(e^x+e^{-x}\right)$$
With that the first term is
$$\frac{1}{\cosh x}=\frac{2}{e^x+e^{-x}}$$
And I think then you can see the limits $\lim_{x\to\pm\infty}\frac{1}{\cosh x}=0$
The second term (neglecting terms of $e^{-x}$ at $x\to\infty$):
$$\lim_{x\to+\infty}\ln\left(\frac{\cosh x}{\cosh x+1}\right)\approx \lim_{x\to+\infty}\ln\left(\frac{e^x}{e^x+2}\right) \approx \lim_{x\to+\infty}\ln\left(\frac{e^x}{e^x}\right) = \lim_{x\to+\infty}\ln 1 = 0$$
For $x\to-\infty$ it is similar.
A: $$
\begin{aligned}
\lim _{x\to \infty }\left(\frac{1}{\cosh \left(x\right)}+\ln\left(\frac{\cosh \left(x\right)}{1+\cosh \left(x\right)}\right)\right)
& = \lim _{x\to \infty }\left(\frac{1+\ln \left(\frac{\left(\frac{e^x+e^{-x}}{2}\right)}{\left(\frac{e^x+e^{-x}}{2}\right)+1}\right)\left(\frac{e^x+e^{-x}}{2}\right)}{\left(\frac{e^x+e^{-x}}{2}\right)}\right)
\\& = \lim _{x\to \infty }\left(\frac{2+\ln \:\left(\frac{e^x+e^{-x}}{e^x+e^{-x}+2}\right)\left(e^x+e^{-x}\right)}{e^x+e^{-x}}\right)
\\& \approx \lim _{x\to \infty }\left(\frac{2+\ln \:\left(\frac{e^x}{e^x+2}\right)\left(e^x\right)}{e^x}\right)
\\& \approx \lim _{x\to \infty }\left(\frac{2+0\cdot \left(e^x\right)}{e^x}\right)
\\& = \color{red}{0}
\end{aligned}
$$
A: For $x\to \pm\infty$, 
$$\lim\limits_{x\to \pm\infty}\underbrace{\log\bigg(\frac{\cos h x}{1+\cos h x}\bigg)}_{\to 0}+\lim\limits_{x\to \pm\infty}\underbrace{\sec h x}_{\to 0}$$
A: 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 $\lim_{x\to\infty}e^x=\infty$
A: We have $$\lim_{x \to \infty}\frac{1}{\cosh(x)}+\log\left(\frac{\cosh(x)}{1+\cosh(x)}\right) = \lim_{x \to \infty}\frac{1}{\cosh(x)}+\log\left(\frac{1}{\frac{1}{\cosh(x)}+1}\right) = 0 + \log(1) = 0$$
since
$$\lim_{x\to \infty} \cosh(x) = \lim_{x\to \infty} {e^x + e^{-x}\over2} = \infty$$
A: For any $x>0$ we have that $1-\frac1x<\log x < x-1$ therefore $$-\frac{1}{\cosh x}<\log \frac{\cosh x}{1+\cosh x}<-\frac{1}{1+\cosh x}$$
From here we obtain $$0<\frac{1}{\cosh x}+\log \frac{\cosh x}{1+\cosh x}<\frac{1}{(1+\cosh x)\cosh x}$$
now bells should ring :-)
