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: Use the definition of hyperbolic cosine.
We know that \[ \cosh x=\frac{e^x+e^{-x}}{2} \] Therefore, \[ \log_e(\cosh x)+x = \log\left(\frac{e^x+e^{-x}}{2}\right)+x \]

Step 2: Simplify the logarithmic expression.
Using logarithmic properties, \[ = \log(e^x+e^{-x})-\log 2+x \] Factor \(e^{-x}\) from the bracket: \[ e^x+e^{-x} = e^{-x}(e^{2x}+1) \] Hence, \[ \log(e^x+e^{-x}) = \log\left(e^{-x}(e^{2x}+1)\right) \] \[ = \log(e^{-x})+\log(e^{2x}+1) \] \[ = -x+\log(e^{2x}+1) \] Thus, \[ \log(\cosh x)+x = -x+\log(e^{2x}+1)-\log 2+x \] \[ = \log(e^{2x}+1)-\log 2 \]

Step 3: Apply the limit.
As \[ x\to -\infty, \] we have \[ e^{2x}\to 0 \] Therefore, \[ \log(e^{2x}+1)\to \log 1=0 \] Hence, \[ \lim_{x\to -\infty}\left[\log(\cosh x)+x\right] = 0-\log 2 \] \[ = -\log 2 \]

Step 4: Final conclusion.
Therefore, \[ \boxed{-\log 2} \]