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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?

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    $\begingroup$ lol, 6 answers in 5 minutes $\endgroup$ – tired Sep 26 '15 at 15:42
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    $\begingroup$ @tired That was almost shocking. $\endgroup$ – Mark Viola Sep 26 '15 at 15:42
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Note that $x-\ln\cosh x=\ln e^x-\ln(e^x+e^{-x})+\ln2=\ln2-\ln(1+e^{-2x})$. Now let $x\to\infty$.

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

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

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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

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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.

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

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HINT:

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

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