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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$-\log 2$

Vedclass · Answer

(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 $\log _e(1 + 0) - \log _e 2 = 0 - \log _e 2 = -\log _e 2$.

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

Similar Questions

$\mathop {\lim }\limits_{x \to 3} [x] = $,(where $[.]$ denotes the greatest integer function)

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 - \cos 2(x - 1)} }}{{x - 1}}$

$\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9} = $

$\mathop {\lim }\limits_{x \to 0} \frac{|x|}{x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module