lim_(x→2) ((((2x+4))^(1/3) −2)/(x^2 −x−2))=?


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Question Number 102034 by Study last updated on 06/Jul/20

$l i \underset{x \to 2}{m} \frac{\sqrt[3]{2 x + 4} - 2}{x^{2} - x - 2} = ?$
	Answered by Dwaipayan Shikari last updated on 06/Jul/20
	$lim_(x→2) ((((2x+4))^(1/3) −2)/(2x+4−8)).((2x−4)/((x−2)(x+1)))=(1/3)8^((−2)/3) .(2/3)=(1/(18)) {(((2x+3)^(1/3) −(8)^(1/3) )/(2x+4−8))=(1/3)8^(−(2/3)) or you can useL hospital$
	$\lim_{x \to 2} \frac{\sqrt[3]{2 x + 4} - 2}{2 x + 4 - 8} . \frac{2 x - 4}{(x - 2) (x + 1)} = \frac{1}{3} 8^{\frac{- 2}{3}} . \frac{2}{3} = \frac{1}{18}$ ${\frac{{(2 x + 3)}^{\frac{1}{3}} - {(8)}^{\frac{1}{3}}}{2 x + 4 - 8} = \frac{1}{3} 8^{- \frac{2}{3}}$ $o r y o u c a n u s e L h o s p i t a l$
	Commented by Study last updated on 06/Jul/20

	$w h a t i s t h e f u r m o l l a h ?$
	Answered by bemath last updated on 06/Jul/20

	$\lim_{x \to 2} \frac{1}{(x + 1)} . \lim_{x \to 2} \frac{\sqrt[3]{2 x + 4} - 2}{x - 2} =$ $\frac{1}{3} . \lim_{x \to 2} \frac{2}{3 \sqrt[3]{{(2 x + 4)}^{2}}} =$ $\frac{2}{9 \times 4} = \frac{1}{18}$
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