let f(x) = ln( 1 + cosh 2x) show that lim_(x→∞) f(x) = 2x − ln 2 hence deduce lim_(x→−∞) f(x) with the asympotes of the curve.


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Question Number 98309 by Rio Michael last updated on 12/Jun/20

$let f (x) = \ln (1 + \cosh 2 x)$ $show that \lim_{x \to \infty} f (x) = 2 x - \ln 2$ $hence deduce \lim_{x \to - \infty} f (x) with the asympotes of the$ $curve .$
	Answered by abdomathmax last updated on 13/Jun/20

	$f (x) = \ln (1 + \frac{e^{2 x} + e^{- 2 x}}{2}) = \ln (2 + e^{2 x} + e^{- 2 x}) - \ln (2)$ $f (x) = \ln (e^{2 x} ({2 e}^{- 2 x} + 1 + e^{- 4 x})) - \ln (2)$ $= 2 x + \ln ({2 e}^{- 2 x} + 1 + e^{- 4 x}) - \ln (2) \Rightarrow$ $\lim_{x \to + \infty} f (x) = \lim_{x \to + \infty} 2 x - \ln (2) = + \infty$ $\lim_{x \to - \infty} f (x) = \lim_{x \to - \infty} \ln (2 + e^{- 2 x}) - \ln (2) = + \infty$ $\lim_{x \to + \infty} \frac{f (x)}{x} = \lim_{x \to + \infty} 2 + \frac{\ln ({2 e}^{- 2 x} + 1 + e^{- 4 x}) - \ln 2}{x}$ $= 2$ $\lim_{x \to + \infty} (f (x) - 2 x) = \lim_{x \to + \infty} \ln ({2 e}^{- 2 x} + 1 + e^{- 4 x}) - \ln 2$ $= - \ln 2 so the asshmptote is the line$ $y = 2 x - \ln (2) at + \infty$
	Commented by mathmax by abdo last updated on 13/Jun/20

	$you are welcome .$
	Commented by Rio Michael last updated on 13/Jun/20

	$brilliant sir . . my textbook was messing things up .$
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