lim_(x→∞) (((3x+6)/(3x+2)))^(x+2) =?


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Question Number 125722 by Mammadli last updated on 13/Dec/20

$\underset{x \to \infty}{l i m} {(\frac{3 x + 6}{3 x + 2})}^{x + 2} = ?$
	Answered by liberty last updated on 13/Dec/20
	$let us denote h(x)=((3x+6)/(3x+2)) ; r(x)=x+2 { ((lim_(x→∞) h(x)=lim_(x→∞) ((3x+6)/(3x+2)) = 1)),((lim_(x→∞) r(x)= lim_(x→∞) (x+2)=∞)) :} use the formula lim_(x→∞) (((3x+6)/(3x+2)))^(x+2) =e^(lim_(x→∞) r(x) [h(x)−1 ]) where h(x)−1 = ((3x+6)/(3x+2))−1=(4/(3x+2)) thus lim_(x→∞) (((3x+6)/(3x+2)))^(x+2) = e^(lim_(x→∞) (x+2)((4/(3x+2)))) = e^(4/3) = (e^4 )^(1/3) .$
	$l e t u s d e n o t e h (x) = \frac{3 x + 6}{3 x + 2}; r (x) = x + 2$ ${\begin{cases} \lim_{x \to \infty} h (x) = \lim_{x \to \infty} \frac{3 x + 6}{3 x + 2} = 1 \\ \lim_{x \to \infty} r (x) = \lim_{x \to \infty} (x + 2) = \infty \end{cases}$ $u s e t h e f o r m u l a \lim_{x \to \infty} {(\frac{3 x + 6}{3 x + 2})}^{x + 2} = e^{\lim_{x \to \infty} r (x) [h (x) - 1]}$ $w h e r e h (x) - 1 = \frac{3 x + 6}{3 x + 2} - 1 = \frac{4}{3 x + 2}$ $t h u s \lim_{x \to \infty} {(\frac{3 x + 6}{3 x + 2})}^{x + 2} = e^{\lim_{x \to \infty} (x + 2) (\frac{4}{3 x + 2})}$ $= e^{\frac{4}{3}} = \sqrt[3]{e^{4}} .$
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