$\epsilon$-proof for the limit $f(x)=\frac{1}{\cosh{(x)}}+\log{\left ( \frac{\cosh{(x)}}{1+\cosh{(x)}} \right )}$ as $x$ goes to $\pm \infty$

Question

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 $x \to +\infty$

$$\lim_{x \ \to \ +\infty} f(x) = L \Leftrightarrow \forall \ \epsilon>0\; (\exists \ \delta : (\;x>\delta\implies |f(x) - L|\leq\epsilon)) $$

2. Limit as $x \to -\infty$

$$\lim_{x \ \to \ -\infty} f(x) = L \Leftrightarrow \forall \ \epsilon>0\; (\exists \ \delta : (\;x<\delta\implies |f(x) - L|\leq\epsilon)) $$

3. Limit evaluating to +$\infty$

$$\lim_{x \ \to \ a} f(x) = +\infty \Leftrightarrow \forall M > 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies f(x) >M)$$

4. Limit evaluating to -$\infty$

$$\lim_{x \ \to \ a} f(x) = -\infty \Leftrightarrow \forall N < 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies f(x) < N)$$

I am struggeling to approach this problem as;

1. 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 "attack" such a problem?

Answer 1

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

$$0<\frac1{\cosh(x)}\le 2e^{-x}$$

Answer 2

With $y:=\operatorname{sech}x>0$ this becomes $f=y-\ln(1+y)\sim\tfrac12y^2$, so the limit is $0$. We wish to prove $\forall\varepsilon>0\exists\delta>0\forall y(y<\delta\implies|f|<\varepsilon)$. Since $0<f<\tfrac12y^2$, we can choose $\delta=\sqrt{2\varepsilon}$. In particular, $\tfrac12y^2-y+\ln(1+y)=\int_0^y\tfrac{z^2}{1+z}dz>0$. The bound $y<\sqrt{2\varepsilon}$ is, of course, equivalent to $|x|>\operatorname{arcosh}\tfrac{1}{\sqrt{2\varepsilon}}$.