Question

Evaluate \[ \lim_{x\to -\infty}\log_e(\cosh x)+x \]

• A. \(\log 2\)
• B. \(-\log 2\)
• C. \(\log\left(\frac{1}{2}\right)+2\)
• D. \(\log\left(\frac{1}{2}\right)-2\)

Hint:

For limits involving hyperbolic functions, first convert them into exponential form using \[ \cosh x=\frac{e^x+e^{-x}}{2} \] and then simplify before applying the limit.

Answer 1

The Correct Option is B

Step 1: Express $\cosh x$ using its definition.
$\cosh x=(e^x+e^{-x})/2$. So $\log_e(\cosh x)+x=\log\left(\frac{e^x+e^{-x}}{2}\right)+x$.
Step 2: Expand using logarithm properties.
$=\log(e^x+e^{-x})-\log 2+x$.
Step 3: Factor $e^{-x}$ from inside the log.
$e^x+e^{-x}=e^{-x}(e^{2x}+1)$, so $\log(e^x+e^{-x})=-x+\log(e^{2x}+1)$.
Step 4: Simplify the combined expression.
$\log_e(\cosh x)+x=-x+\log(e^{2x}+1)-\log 2+x=\log(e^{2x}+1)-\log 2$.
Step 5: Take the limit as $x\to -\infty$.
As $x\to-\infty$, $e^{2x}\to 0$, so $\log(e^{2x}+1)\to\log 1=0$. \[\lim_{x\to-\infty}[\log_e(\cosh x)+x]=0-\log 2=-\log 2.\]
Step 6: State the answer.
\[ \boxed{-\log 2} \]