lim _{x rightarrow infty} [x - log (cosh x)] | vedclass

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Question

(D) આપણે જાણીએ છીએ કે $\cosh x = \frac{e^x + e^{-x}}{2}$.આ કિંમત લક્ષમાં મૂકતા,આપણને મળે છે:$\lim _{x \rightarrow \infty} [x - \log (\frac{e^x + e^{-x

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(D) આપણે જાણીએ છીએ કે $\cosh x = \frac{e^x + e^{-x}}{2}$.આ કિંમત લક્ષમાં મૂકતા,આપણને મળે છે:$\lim _{x \rightarrow \infty} [x - \log (\frac{e^x + e^{-x}}{2})]$$= \lim _{x \rightarrow \infty} [x - (\log (e^x + e^{-x}) - \log 2)]$$= \lim _{x \rightarrow \infty} [x - (\log e^x + \log (1 + e^{-2x})) + \log 2]$$= \lim _{x \rightarrow \infty} [x - x - \log (1 + e^{-2x}) + \log 2]$$= \lim _{x \rightarrow \infty} [\log 2 - \log (1 + e^{-2x})]$જેમ $x \rightarrow \infty$,તેમ $e^{-2x} \rightarrow 0$,તેથી $\log (1 + e^{-2x}) \rightarrow \log 1 = 0$.તેથી,લક્ષ $\log 2 - 0 = \log 2$ છે.

$\lim _{x \rightarrow \infty} [x - \log (\cosh x)] = $

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