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

Consider the following statements:
$I$. $\lim _{n \rightarrow \infty} \frac{2^n+(-2)^n}{2^n}$ does not exist.
$II$. $\lim _{n \rightarrow \infty} \frac{3^n+(-3)^n}{4^n}$ does not exist.
Then,

The value of the limit $\lim_{x \rightarrow 1} \frac{\sin(e^{x-1}-1)}{\log x}$ is

For each $t \in R$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 0^+} x \left( [\frac{1}{x}] + [\frac{2}{x}] + \dots + [\frac{15}{x}] \right) = $

$\lim \limits_{x}$ ${\rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\cos ^{-1} x\right)-x}{1-\tan \left(\cos ^{-1} x\right)}$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^3} + 1}} + \frac{4}{{{n^3} + 1}} + \frac{9}{{{n^3} + 1}} + \dots + \frac{{{n^2}}}{{{n^3} + 1}}} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module