$\epsilon$-proof for the limit $f(x)=\frac{1}{\cosh{(x)}}+\log{\left ( \frac{\cosh{(x)}}{1+\cosh{(x)}} \right )}$ as $x$ goes to $\pm \infty$ I want to use a $\epsilon,\delta$-proof for the existence and value for the limit of $$f(x)=\frac{1}{\cosh{(x)}}+\log{\left ( \frac{\cosh{(x)}}{1+\cosh{(x)}} \right )}$$
for $x \rightarrow \pm\infty$.
Now, I know the definition for such proof to be:
1. Limit as $x \to +\infty$
$$\lim_{x \ \to \  +\infty} f(x) = L \Leftrightarrow 
  \forall \ \epsilon&gt;0\; (\exists \ \delta : (\;x&gt;\delta\implies |f(x) - L|\leq\epsilon))
$$
2. Limit as $x \to -\infty$
$$\lim_{x \ \to \  -\infty} f(x) = L \Leftrightarrow 
  \forall \ \epsilon&gt;0\; (\exists \ \delta : (\;x&lt;\delta\implies |f(x) - L|\leq\epsilon))
$$
3. Limit evaluating to +$\infty$
$$\lim_{x \ \to \  a} f(x) = +\infty \Leftrightarrow \forall M  &gt; 0 \ (\exists \  \delta &gt; 0  \ : \ (0&lt;|x-a|&lt;\delta \implies f(x) &gt;M)$$
4. Limit evaluating to -$\infty$
$$\lim_{x \ \to \  a} f(x) = -\infty \Leftrightarrow \forall  N  &lt; 0 \ (\exists \  \delta &gt; 0  \ : \ (0&lt;|x-a|&lt;\delta \implies f(x)  &lt; N)$$
I am struggeling to approach this problem as;

Wouldn't I need to know what the limit is before doing the actual proof and thus not being able to know which method to use?

This one seems rather complex and I am completly lost. Can someone explain to me how to identify and &quot;attack&quot; such a problem?
With $y:=\operatorname{sech}x&gt;0$ this becomes $f=y-\ln(1+y)\sim\tfrac12y^2$, so the limit is $0$. We wish to prove $\forall\varepsilon&gt;0\exists\delta&gt;0\forall y(y&lt;\delta\implies|f|&lt;\varepsilon)$. Since $0&lt;f&lt;\tfrac12y^2$, we can choose $\delta=\sqrt{2\varepsilon}$. In particular, $\tfrac12y^2-y+\ln(1+y)=\int_0^y\tfrac{z^2}{1+z}dz&gt;0$. The bound $y&lt;\sqrt{2\varepsilon}$ is, of course, equivalent to $|x|&gt;\operatorname{arcosh}\tfrac{1}{\sqrt{2\varepsilon}}$.
HINTS:
Note that the hyperbolic cosine function is even and so we need only examine the case $x\to +\infty$.
Then, apply the inequalities
$$\frac{y}{1+y}\le \log(y)\le y-1$$
and
$$0&lt;\frac1{\cosh(x)}\le 2e^{-x}$$