find limit when x is infinity

Question

How can we find the following limit. $$\lim_{x\to\infty}(x-\ln\cosh x)$$ where $$\cosh t=\frac{e^t+e^{-t}}{2}.$$

I thought about it alot but didn't get any start

Answer 1

We can write our limit as , $$\ln(\lim_{x\to\infty}e^{x-\ln\cosh x})$$ $$=\ln (\lim_{x\to\infty}\dfrac{e^x}{\cosh x})$$ $$=\ln (\lim_{x\to\infty}\dfrac{2e^x}{e^x+e^{-x}})$$ $$=\ln (\lim_{x\to\infty}\dfrac{2}{1+e^{-2x}})$$ $$=\ln 2$$

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

Answer 2

Note that we have

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

Hence, we can write

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