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Question Number 95980 by john santu last updated on 29/May/20

If lim_(x→3) (((√(3x+7))−((20x+4))^(1/(3 )) +ax+b)/((x−3)^2 )) exist , what is the value of ab

$$\mathrm{If}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3x}+\mathrm{7}}−\sqrt[{\mathrm{3}\:\:}]{\mathrm{20x}+\mathrm{4}}\:+{a}\mathrm{x}+{b}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{exist}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{ab}\: \\ $$

Answered by i jagooll last updated on 29/May/20

set x−3 = y, x=y+3 , y→0 lim_(y→0) (((√(3y+16))−((20y+64))^(1/(3 )) +ay+3a+b)/y^2 ) numerator must be = 0 ⇒4−4+3a+b = 0 , b = −3a L′Hopital lim_(y→0) (((3/(2(√(3y+16))))−((20)/(3 (((20y+64)^2 ))^(1/(3 )) ))+a)/(2y)) ⇒numerator must be = 0 ⇒(3/8)−((20)/(3.16)) +a = 0 a = (5/(12))−(3/8) = ((10−9)/(24)) = (1/(24)) then b = −3a = −(1/8) so a×b = − (1/(192)) .•

$$\mathrm{set}\:{x}−\mathrm{3}\:=\:{y},\:{x}={y}+\mathrm{3}\:,\:{y}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}{y}+\mathrm{16}}−\sqrt[{\mathrm{3}\:\:}]{\mathrm{20}{y}+\mathrm{64}}+{ay}+\mathrm{3}{a}+{b}}{{y}^{\mathrm{2}} } \\ $$$${numerator}\:{must}\:{be}\:=\:\mathrm{0}\: \\ $$$$\Rightarrow\mathrm{4}−\mathrm{4}+\mathrm{3}{a}+{b}\:=\:\mathrm{0}\:,\:{b}\:=\:−\mathrm{3}{a} \\ $$$${L}'\mathrm{Hopital} \\ $$$$\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{3y}+\mathrm{16}}}−\frac{\mathrm{20}}{\mathrm{3}\:\sqrt[{\mathrm{3}\:\:}]{\left(\mathrm{20y}+\mathrm{64}\right)^{\mathrm{2}} }}+{a}}{\mathrm{2}{y}} \\ $$$$\Rightarrow\mathrm{numerator}\:\mathrm{must}\:\mathrm{be}\:=\:\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{20}}{\mathrm{3}.\mathrm{16}}\:+{a}\:=\:\mathrm{0} \\ $$$${a}\:=\:\frac{\mathrm{5}}{\mathrm{12}}−\frac{\mathrm{3}}{\mathrm{8}}\:=\:\frac{\mathrm{10}−\mathrm{9}}{\mathrm{24}}\:=\:\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${then}\:{b}\:=\:−\mathrm{3}{a}\:=\:−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${so}\:{a}×{b}\:=\:−\:\frac{\mathrm{1}}{\mathrm{192}}\:.\bullet \\ $$

Commented by john santu last updated on 29/May/20

greatt

$$\mathrm{greatt} \\ $$

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