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Question Number 39067 by bshahid010@gmail.com last updated on 02/Jul/18

Commented by math khazana by abdo last updated on 03/Jul/18

$= {l i m}_{h \to 0} \frac{2 h - h^{3} \sqrt{8 + h}}{2 h^{4} \sqrt{8 + h}}$ $= {l i m}_{h \to 0} \frac{2 - h^{2} \sqrt{8 + h}}{2 h^{3} \sqrt{8 + h}} s o i f h \to 0^{+}$ $l i m (. . .) = + \infty i f h \to 0^{-}$ $l i m (. . .) = - \infty .$
Commented by bshahid010@gmail.com last updated on 03/Jul/18

$wrong answer$
Commented by math khazana by abdo last updated on 03/Jul/18

$p o s t t h e a n s w e r s i r .$
Commented by bshahid010@gmail.com last updated on 03/Jul/18

$i dont know that answer but given options$ $a r e a) - \frac{1}{12} b) \frac{- 4}{3} c) - \frac{16}{3} d) - \frac{1}{48}$
Commented by math khazana by abdo last updated on 03/Jul/18

$l e t A (h) = \frac{1}{h^{3} \sqrt{8 + h}} - \frac{1}{2 h} (h \neq 0)$ $A (h) = \frac{2 h - h^{3} \sqrt{8 + h}}{2 h^{4} \sqrt{8 + h}} = \frac{2 - h^{2} \sqrt{8 + h}}{2 h^{3} \sqrt{8 + h}}$ $w e h a v e \sqrt{8 + h} = 2 \sqrt{2} \sqrt{1 + \frac{h}{8}} \sim 2 \sqrt{2} (1 + \frac{h}{16}) (h \to 0)$ $2 - h^{2} \sqrt{8 + h} \sim 2 - 2 \sqrt{2} h^{2} - \frac{h^{3}}{16}$ $2 h^{3} \sqrt{8 + h} \sim 4 \sqrt{2} h^{3} (1 + \frac{h}{16}) = 4 \sqrt{2} h^{3} + \frac{\sqrt{2}}{4} h^{4} \Rightarrow$ ${l i m}_{h \to 0} A (h) = {l i m}_{h \to 0} \frac{2}{4 \sqrt{2} h^{3}} = \bar{+} \infty .$
Commented by math khazana by abdo last updated on 03/Jul/18

$f o r m e i h a v e c o n s i d e r e d h^{3} (\sqrt{8 + h}) n o t$ $h (^{3} \sqrt{8 + h}) . . d i d y o u l o o k t h e d i f f e r e n c e ? . . .$
Commented by math khazana by abdo last updated on 03/Jul/18

$m y a n s w e r i s n o t w r o n g b e c a u s e i h a v e t a k e$ $h^{3} (\sqrt{8 + h}) n o t h (^{3} \sqrt{8 + h}) . . . .$
	Answered by ajfour last updated on 03/Jul/18
	$lim_(h→0) {(1/(h ((8+h))^(1/3) ))−(1/(2h))} = lim_(h→0) ((1−(h/(24))−1)/(2h)) = −(1/(48)) .$
	$\lim_{h \to 0} {\frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2 h}}$ $= \lim_{h \to 0} \frac{1 - \frac{h}{24} - 1}{2 h} = - \frac{1}{48} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18
	$lim_(h→0) (1/h){(1/((8+h)^(1/3) ))−(1/2)} lim_(h→0) (1/h){((2−(8+h)^(1/3) )/(2(8+h)^(1/3) ))} lim_(h→0) (1/h){(((8)^(1/3) −(8+h)^(1/3) )/(2(8+h)^(1/3) ))} lim_(h→0) (1/h){((8−8−h)/(2(8+h)^(1/3) {(8)^(2/3) +8^(1/3) (8+h)^(1/3) +(8+h)^(2/3) })) a^3 −b^3 =(a−b)(a^2 +ab+b^2 ) =((−1)/(2×(8)^(1/3) {3×8^(2/3) })) =((−1)/(6×8^((1/3)+(2/3)) ))=((−1)/(6×8^1 ))=((−1)/(48))$
	$\lim_{h \to 0} \frac{1}{h} {\frac{1}{{(8 + h)}^{\frac{1}{3}}} - \frac{1}{2}}$ $\lim_{h \to 0} \frac{1}{h} {\frac{2 - {(8 + h)}^{\frac{1}{3}}}{2 {(8 + h)}^{\frac{1}{3}}}}$ $\lim_{h \to 0} \frac{1}{h} {\frac{{(8)}^{\frac{1}{3}} - {(8 + h)}^{\frac{1}{3}}}{2 {(8 + h)}^{\frac{1}{3}}}}$ $\lim_{h \to 0} \frac{1}{h} {\frac{8 - 8 - h}{2 {(8 + h)}^{\frac{1}{3}} {{(8)}^{\frac{2}{3}} + 8^{\frac{1}{3}} {(8 + h)}^{\frac{1}{3}} + {(8 + h)}^{\frac{2}{3}}}}$ $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $= \frac{- 1}{2 \times {(8)}^{\frac{1}{3}} {3 \times 8^{\frac{2}{3}}}}$ $= \frac{- 1}{6 \times 8^{\frac{1}{3} + \frac{2}{3}}} = \frac{- 1}{6 \times 8^{1}} = \frac{- 1}{48}$
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