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?

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