lim _{x rightarrow infty} [x - log (cosh x)] | vedclass

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(D) We know that $\cosh x = \frac{e^x + e^{-x}}{2}$.Substituting this into the limit,we get:$\lim _{x \rightarrow \infty} [x - \log (\frac{e^x + e^{-x

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(D) We know that $\cosh x = \frac{e^x + e^{-x}}{2}$.Substituting this into the limit,we get:$\lim _{x \rightarrow \infty} [x - \log (\frac{e^x + e^{-x}}{2})]$$= \lim _{x \rightarrow \infty} [x - (\log (e^x + e^{-x}) - \log 2)]$$= \lim _{x \rightarrow \infty} [x - \log (e^x(1 + e^{-2x})) + \log 2]$$= \lim _{x \rightarrow \infty} [x - (\log e^x + \log (1 + e^{-2x})) + \log 2]$$= \lim _{x \rightarrow \infty} [x - x - \log (1 + e^{-2x}) + \log 2]$$= \lim _{x \rightarrow \infty} [\log 2 - \log (1 + e^{-2x})]$As $x \rightarrow \infty$,$e^{-2x} \rightarrow 0$,so $\log (1 + e^{-2x}) \rightarrow \log 1 = 0$.Therefore,the limit is $\log 2 - 0 = \log 2$.

$\lim _{x \rightarrow \infty} [x - \log (\cosh x)] = $

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