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Question-39067

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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
=lim_(h→0) ((2h −h^3 (√(8+h)))/(2h^4 (√(8+h)))) =lim_(h→0) ((2 −h^2 (√(8+h)))/(2h^3 (√(8+h)))) so if h→0^+ lim(...) = +∞ if h→0^− lim(...)= −∞ .
$$={lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{2}{h}\:−{h}^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}}{\mathrm{2}{h}^{\mathrm{4}} \sqrt{\mathrm{8}+{h}}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{2}\:−{h}^{\mathrm{2}} \sqrt{\mathrm{8}+{h}}}{\mathrm{2}{h}^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}}\:\:{so}\:{if}\:{h}\rightarrow\mathrm{0}^{+} \\ $$$${lim}\left(…\right)\:=\:+\infty\:\:\:\:{if}\:{h}\rightarrow\mathrm{0}^{−} \\ $$$${lim}\left(…\right)=\:−\infty\:. \\ $$
Commented by bshahid010@gmail.com last updated on 03/Jul/18
wrong answer
$$\:\mathrm{wrong}\:\mathrm{answer} \\ $$
Commented by math khazana by abdo last updated on 03/Jul/18
post the answer sir.
$${post}\:{the}\:{answer}\:{sir}. \\ $$
Commented by bshahid010@gmail.com last updated on 03/Jul/18
i dont know that answer but given options are a)−(1/(12)) b)((−4)/3) c)−((16)/3) d)−(1/(48))
$$\mathrm{i}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{that}\:\mathrm{answer}\:\mathrm{but}\:\mathrm{given}\:\mathrm{options} \\ $$$$\left.\mathrm{a}\left.\mathrm{r}\left.\mathrm{e}\left.\:\mathrm{a}\right)−\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{b}\right)\frac{−\mathrm{4}}{\mathrm{3}}\:\mathrm{c}\right)−\frac{\mathrm{16}}{\mathrm{3}}\:\mathrm{d}\right)−\frac{\mathrm{1}}{\mathrm{48}} \\ $$
Commented by math khazana by abdo last updated on 03/Jul/18
let A(h) = (1/(h^3 (√(8+h)))) −(1/(2h)) (h≠0) A(h) = ((2h −h^3 (√(8+h)))/(2h^4 (√(8+h)))) = ((2−h^2 (√(8+h)))/(2h^3 (√(8+h)))) we have (√(8+h))=2(√2)(√(1+(h/8))) ∼2(√2)(1+(h/(16))) (h→0) 2−h^2 (√(8+h)) ∼2−2(√2)h^2 −(h^3 /(16)) 2h^3 (√(8+h)) ∼ 4(√2)h^3 (1+(h/(16)) )=4(√2)h^3 +((√2)/4) h^4 ⇒ lim_(h→0) A(h) =lim_(h→0) (2/(4(√2)h^3 )) =+^− ∞ .
$${let}\:\:{A}\left({h}\right)\:=\:\frac{\mathrm{1}}{{h}^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}}\:−\frac{\mathrm{1}}{\mathrm{2}{h}}\:\:\:\left({h}\neq\mathrm{0}\right) \\ $$$${A}\left({h}\right)\:=\:\frac{\mathrm{2}{h}\:−{h}^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}}{\mathrm{2}{h}^{\mathrm{4}} \sqrt{\mathrm{8}+{h}}}\:=\:\frac{\mathrm{2}−{h}^{\mathrm{2}} \sqrt{\mathrm{8}+{h}}}{\mathrm{2}{h}^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}} \\ $$$${we}\:{have}\:\sqrt{\mathrm{8}+{h}}=\mathrm{2}\sqrt{\mathrm{2}}\sqrt{\mathrm{1}+\frac{{h}}{\mathrm{8}}}\:\:\sim\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}+\frac{{h}}{\mathrm{16}}\right)\:\left({h}\rightarrow\mathrm{0}\right) \\ $$$$\mathrm{2}−{h}^{\mathrm{2}} \sqrt{\mathrm{8}+{h}}\:\sim\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}{h}^{\mathrm{2}} −\frac{{h}^{\mathrm{3}} }{\mathrm{16}} \\ $$$$\mathrm{2}{h}^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}\:\sim\:\mathrm{4}\sqrt{\mathrm{2}}{h}^{\mathrm{3}} \left(\mathrm{1}+\frac{{h}}{\mathrm{16}}\:\right)=\mathrm{4}\sqrt{\mathrm{2}}{h}^{\mathrm{3}} \:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:{h}^{\mathrm{4}} \:\Rightarrow \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} {A}\left({h}\right)\:={lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}{h}^{\mathrm{3}} }\:\:=\overset{−} {+}\:\infty\:. \\ $$
Commented by math khazana by abdo last updated on 03/Jul/18
for me i have considered h^3 ((√(8+h))) not h(^3 (√(8+h)))..did you look the difference ?...
$${for}\:{me}\:{i}\:{have}\:{considered}\:{h}^{\mathrm{3}} \left(\sqrt{\mathrm{8}+{h}}\right)\:{not} \\ $$$${h}\left(^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}\right)..{did}\:{you}\:{look}\:{the}\:{difference}\:?… \\ $$
Commented by math khazana by abdo last updated on 03/Jul/18
my answer is not wrong because i have take h^3 ((√(8+h))) not h(^3 (√(8+h)))....
$${my}\:{answer}\:{is}\:{not}\:{wrong}\:{because}\:{i}\:{have}\:{take} \\ $$$${h}^{\mathrm{3}} \left(\sqrt{\mathrm{8}+{h}}\right)\:{not}\:{h}\left(^{\mathrm{3}} \sqrt{\mathrm{8}+{h}}\right)…. \\ $$
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)) .
$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{h}\:\sqrt[{\mathrm{3}}]{\mathrm{8}+{h}}}−\frac{\mathrm{1}}{\mathrm{2}{h}}\right\} \\ $$$$\:\:\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\frac{{h}}{\mathrm{24}}−\mathrm{1}}{\mathrm{2}{h}}\:=\:−\frac{\mathrm{1}}{\mathrm{48}}\:. \\ $$
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))
$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{h}}\left\{\frac{\mathrm{1}}{\left(\mathrm{8}+{h}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{h}}\left\{\frac{\mathrm{2}−\left(\mathrm{8}+{h}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{2}\left(\mathrm{8}+{h}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }\right\} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{h}}\left\{\frac{\left(\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\mathrm{8}+{h}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{2}\left(\mathrm{8}+{h}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }\right\} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{h}}\left\{\frac{\mathrm{8}−\mathrm{8}−{h}}{\mathrm{2}\left(\mathrm{8}+{h}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left\{\left(\mathrm{8}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{8}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{8}+{h}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\left(\mathrm{8}+{h}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \right\}}\right. \\ $$$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}×\left(\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left\{\mathrm{3}×\mathrm{8}^{\frac{\mathrm{2}}{\mathrm{3}}} \right\}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{6}×\mathrm{8}^{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{−\mathrm{1}}{\mathrm{6}×\mathrm{8}^{\mathrm{1}} }=\frac{−\mathrm{1}}{\mathrm{48}} \\ $$

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