# Prove that $\lim_{x\to \infty}\left(x-\ln\cosh x\right)=\ln 2$

Prove that $\lim_{x\to \infty}\left(x-\ln\cosh x\right)=\ln 2$

I used L Hospital Rule but it does not simplify.Then i expanded $\cosh x$ by using McLaurin series but due to $x\to \infty$,this is also not working.How should i evaluate this limit?

• lol, 6 answers in 5 minutes – tired Sep 26 '15 at 15:42
• @tired That was almost shocking. – Mark Viola Sep 26 '15 at 15:42

Note that $x-\ln\cosh x=\ln e^x-\ln(e^x+e^{-x})+\ln2=\ln2-\ln(1+e^{-2x})$. Now let $x\to\infty$.

$$\lim_{x\to\infty}e^{x-\ln\cosh x}$$ $$=\lim_{x\to\infty}\dfrac{e^x}{\cosh x}$$ $$=\lim_{x\to\infty}\dfrac{2e^x}{e^x+e^{-x}}$$ $$=\lim_{x\to\infty}\dfrac{2}{1+e^{-2x}}$$ $$=2$$

Thus, $\lim_{x\to\infty}(x-\ln\cosh x)=\ln 2$

For $x\rightarrow \infty$, $\cosh(x)\sim \frac{e^{x}}{2}$, Therefore:

$$\log(\cosh(x))\sim x-\log(2)$$ and $$\lim_{x\rightarrow\infty}[x-\log(\cosh(x))]=\log(2)$$

QED

Let $y=x-\ln \cosh(x)$. Consider $e^y$. We have \begin{align*} \lim_{x\to \infty} e^y&= \lim_{x\to \infty} \frac{e^x}{e^{\ln \cosh(x)}}\\ &= \lim_{x\to \infty} \frac{e^x}{\cosh(x)}\\ &= \lim_{x\to \infty} \frac{2e^x}{e^x + e^{-x}}\\ &= \lim_{x\to \infty} \frac{2}{1 + e^{-2x}}\\ &= 2. \end{align*}

Taking the natural log, we get what we want.

Try writing $$x = log (e^x)$$ And then: $$x-log(coshx) = log(e^x) - log (\dfrac{e^x+e^{-x}}{2}) = -log(\dfrac{1+e^{-2x}}{2})$$

HINT:

$\log \cosh x=\log\left(\frac{e^x+e^{-x}}{2}\right)=x-\log 2+\log(1+e^{-2x})$