lim_(x→∞) (((x+a)/(x−a)))^x ?


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Question Number 111442 by bobhans last updated on 03/Sep/20

$\lim_{x \to \infty} {(\frac{x + a}{x - a})}^{x} ?$
	Answered by bemath last updated on 03/Sep/20

	Answered by ajfour last updated on 03/Sep/20
	$ln L=lim_(x→∞) xln (((1+(a/x))/(1−(a/x)))) = lim_(x→∞) {xln (1+(a/x))−xln (1−(a/x))} = alim_(x→∞) {((ln (1+(a/x)))/(((a/x))))+((ln (1−(a/x)))/((−(a/x))))} ln L = 2a L = e^(2a) .$
	$\ln L = \lim_{x \to \infty} x \ln (\frac{1 + \frac{a}{x}}{1 - \frac{a}{x}})$ $= \lim_{x \to \infty} {x \ln (1 + \frac{a}{x}) - x \ln (1 - \frac{a}{x})}$ $= a \lim_{x \to \infty} {\frac{\ln (1 + \frac{a}{x})}{(\frac{a}{x})} + \frac{\ln (1 - \frac{a}{x})}{(- \frac{a}{x})}}$ $\ln L = 2 a$ $L = e^{2 a} .$
	Answered by Dwaipayan Shikari last updated on 03/Sep/20

	$\lim_{x \to \infty} {(1 + \frac{2 a}{x - a})}^{\frac{x}{2 a} . \frac{x - a}{x - a} . 2 a}$ $= e^{\frac{2 a x}{x - a}} = e^{2 a}$
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