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?
 A: $$\lim_{x\to\infty}e^{x-\ln\cosh x}$$
$$=\lim_{x\to\infty}\dfrac{e^x}{\cosh x}$$
$$=\lim_{x\to\infty}\dfrac{2e^x}{e^x+e^{-x}}$$
$$=\lim_{x\to\infty}\dfrac{2}{1+e^{-2x}}$$
$$=2$$
Thus, $\lim_{x\to\infty}(x-\ln\cosh x)=\ln 2$
A: Note that $x-\ln\cosh x=\ln e^x-\ln(e^x+e^{-x})+\ln2=\ln2-\ln(1+e^{-2x})$. Now let $x\to\infty$.
A: 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
A: 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.
A: Try writing $$x = log (e^x)$$
And then:
$$x-log(coshx) = log(e^x) - log (\dfrac{e^x+e^{-x}}{2}) = -log(\dfrac{1+e^{-2x}}{2}) $$
A: HINT:
$\log \cosh x=\log\left(\frac{e^x+e^{-x}}{2}\right)=x-\log 2+\log(1+e^{-2x})$
