Limit of function of hyperbolic

Question

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.

Answer 1

For $x\to \pm\infty$,

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

Answer 2

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.

Answer 3

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 $\lim_{x\to\infty}e^x=\infty$

Answer 4

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

$$\lim_{x\to \infty} \cosh(x) = \lim_{x\to \infty} {e^x + e^{-x}\over2} = \infty$$

Answer 5

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 :-)