lim _{x rightarrow-infty} log _e(cosh x)+x= | vedclass

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

Vedclass · Accepted Answer

$-\log 2$

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(B) We know that $\cosh x = \frac{e^x + e^{-x}}{2}$.Substituting this into the limit,we get $\lim _{x \rightarrow-\infty} \log _e \left( \frac{e^x + e^{-x}}{2} \right) + x$.Using the property $\log(a/b) = \log a - \log b$,we have $\lim _{x \rightarrow-\infty} [\log _e(e^x + e^{-x}) - \log _e 2 + x]$.Since $x \rightarrow -\infty$,$e^x \rightarrow 0$. We can factor out $e^{-x}$ inside the log: $\log _e(e^{-x}(e^{2x} + 1)) = \log _e(e^{-x}) + \log _e(1 + e^{2x}) = -x + \log _e(1 + e^{2x})$.Substituting this back: $\lim _{x \rightarrow-\infty} [-x + \log _e(1 + e^{2x}) - \log _e 2 + x]$.The terms $x$ and $-x$ cancel out,leaving $\lim _{x \rightarrow-\infty} [\log _e(1 + e^{2x}) - \log _e 2]$.As $x \rightarrow -\infty$,$e^{2x} \rightarrow 0$,so $\log _e(1 + 0) - \log _e 2 = 0 - \log _e 2 = -\log _e 2$.

$\lim _{x \rightarrow-\infty} \log _e(\cosh x)+x=$

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