I am hoping you guys can help me, since it seems that i'm doing something wrong.
This task is to be solved without the use of differential calculus.
I have the function:
$f(x)= \frac{1}{\cosh x} + \log (\frac{\cosh x}{1 + \cosh x})$
And the task is to show what happens when $x \to \pm \; \infty$. I also have to show that $f(x) \geq 0$ for all $x \in R$.
Here is my approach:
$x \to \infty$:
$\cosh x$ approaches infity so $\frac{1}{\cosh x}$ approaches 0.
$\log (\frac{\cosh x}{1 + \cosh x})$ here the + 1 matters less and less so it approaches log(1)=0. Therefore $\lim_{x\to\infty} f(x) = 0$.
The approach is the same as for $x \to - \infty$ because cosh x is reflective.
I have already proved in an earlier task that $cosh x \geq 1$ for all x so i'm thinking i could use that to say something about logarithm and the $\frac{1}{\cosh x}$ Cancelling out or something, but i have no idea how to show it mathematically.
I am hoping you guys can help me. Thanks in advance.
Note that $$ \log\left(\frac{\cosh{x}}{1+\cosh{x}}\right)=-\log\left(\frac{1+\cosh{x}}{\cosh{x}}\right)=-\log\left(1+\frac{1}{\cosh{x}}\right). $$ This is a nice trick to use whenever you have an annoying quantity inside of a $\log$. Now, we have the standard bound $\log{r}\leq (r-1)$, which is valid for all $r>0$. You said that you showed before that $\cosh{x}\geq 1$ for all $x\in\mathbb{R}$, so certainly $1+\frac{1}{\cosh{x}}>0$ and we can apply the bound. This implies that $$ \log\left(1+\frac{1}{\cosh{x}}\right)\leq \frac{1}{\cosh{x}}, $$ and hence that $$ -\log\left(1+\frac{1}{\cosh{x}}\right)\geq -\frac{1}{\cosh{x}}. $$ Thus, $$ \frac{1}{\cosh{x}}+\log\left(\frac{\cosh{x}}{1+\cosh{x}}\right)\geq \frac{1}{\cosh{x}}-\frac{1}{\cosh{x}}=0 $$ for all $x\in\mathbb{R}$.