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$\lim _{x \rightarrow-\infty} \log _e(\cosh x)+x=$
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Verified Answer
The correct answer is:
$-\log 2$
Here, $\lim _{x \rightarrow-\infty} \log _e(\cosh x)+x$
$=\lim _{x \rightarrow-\infty} \log \left(e^x+e^{-x}\right)+\ln 2$
$\Rightarrow \quad \lim _{x \rightarrow-\infty} \log \left(1+e^{2 x}\right)+\ln 2$
$=-\ln 2$
$=\lim _{x \rightarrow-\infty} \log \left(e^x+e^{-x}\right)+\ln 2$
$\Rightarrow \quad \lim _{x \rightarrow-\infty} \log \left(1+e^{2 x}\right)+\ln 2$
$=-\ln 2$
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