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Question Number 142333 by mohammad17 last updated on 30/May/21

Commented by mohammad17 last updated on 30/May/21

$p l e a s e s i r h e l p m e$
	Answered by Dwaipayan Shikari last updated on 30/May/21

	$\lim_{x \to 0} \frac{{(x + 4)}^{3 / 2} - 8}{x} = \frac{8 ({(1 + \frac{x}{4})}^{3 / 2} - 1)}{x} = 8 \frac{1 + \frac{x}{4} . \frac{3}{2} - 1}{x} = 3$
	Commented by mathmax by abdo last updated on 30/May/21
	$let f(x)=(((x+4)^(3/2) −8)/x) ⇒f(x)=((4^(3/2) (1+(x/4))^(3/2) −8)/x) =((8{(1+(x/4))^(3/2) −1})/x)∼((8(1+((3x)/8)−1))/x)=3 ⇒lim_(x→0) f(x)=3$
	$let f (x) = \frac{{(x + 4)}^{\frac{3}{2}} - 8}{x} \Rightarrow f (x) = \frac{4^{\frac{3}{2}} {(1 + \frac{x}{4})}^{\frac{3}{2}} - 8}{x}$ $= \frac{8 {{(1 + \frac{x}{4})}^{\frac{3}{2}} - 1}}{x} \sim \frac{8 (1 + \frac{3 x}{8} - 1)}{x} = 3 \Rightarrow \lim_{x \to 0} f (x) = 3$
	Answered by bramlexs22 last updated on 30/May/21

	$f (x) = \lim_{x \to 0} \frac{\sqrt{{(x + 4)}^{3}} - 8}{x}$ $f (x) = \lim_{x \to 0} \frac{{(x + 4)}^{3} - 64}{x} . \lim_{x \to 0} \frac{1}{\sqrt{{(x + 4)}^{3}} + 8}$ $= \lim_{x \to 0} \frac{x (x^{2} + 12 x + 48)}{x} . \frac{1}{16}$ $= \frac{48}{16} = 3$
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