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Question Number 211895 by Spillover last updated on 23/Sep/24

	Answered by mehdee7396 last updated on 23/Sep/24

	${l i m}_{x \to \infty} (1 + \frac{a}{x} - \frac{4}{x^{2}} - 1) 2 x$ $= {l i m}_{x \to \infty} (\frac{a x - 4}{x^{2}}) 2 x = 2 a$ $\Rightarrow {l i m}_{x \to \infty} {(1 + \frac{a}{x} - \frac{4}{x^{2}} - 1)}^{2 x} = e^{2 a}$ $\Rightarrow 2 a = 3 \Rightarrow a = \frac{3}{2} ✓$
	Answered by BHOOPENDRA last updated on 23/Sep/24
	$we know if we have function lim_(x→a) f(x)^(g(x)) =1^∞ then we use the property lim_(e^x →a) g(x) {f(x)−1} So lim_(e^x →∞) 2x (1+(a/x)−(4/x^2 ) −1)=e^3 lim_(e^x →∞) 2x ((a/x)−(4/x^2 ))=e^3 e^(2(a−(4/∞))) =e^3 e^(2a) =e^3 2a=3 a=(3/2)$
	$w e k n o w i f w e h a v e f u n c t i o n$ $\lim_{x \to a} f {(x)}^{g (x)} = 1^{\infty}$ $t h e n w e u s e t h e p r o p e r t y$ $\lim_{e^{x} \to a} g (x) {f (x) - 1}$ $S o \lim_{e^{x} \to \infty} 2 x (1 + \frac{a}{x} - \frac{4}{x^{2}} - 1) = e^{3}$ $\lim_{e^{x} \to \infty} 2 x (\frac{a}{x} - \frac{4}{x^{2}}) = e^{3}$ $e^{2 (a - \frac{4}{\infty})} = e^{3}$ $e^{2 a} = e^{3}$ $2 a = 3$ $a = \frac{3}{2}$
	Commented by Spillover last updated on 24/Sep/24

	$t h a n k s$
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