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without-L-hopital-lim-x-1-2-2x-3-3x-2-a-bx-4x-2-1-3-4-find-a-b-

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Question Number 84568 by jagoll last updated on 14/Mar/20
without L′hopital lim_(x→−(1/2)) ((2x^3 +3x^2 −(√(a+bx)))/(4x^2 −1)) = −(3/4) find a+b
$$\mathrm{without}\:\mathrm{L}'\mathrm{hopital} \\ $$$$\underset{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{2x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\sqrt{\mathrm{a}+\mathrm{bx}}}{\mathrm{4x}^{\mathrm{2}} −\mathrm{1}}\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b} \\ $$
Commented by john santu last updated on 14/Mar/20
(1) 2(−(1/8))+3((1/4))−(√(a−(b/2))) = 0 −(1/4)+(3/4) = (√(a−(b/2))) ⇒ a= ((1+2b)/4) (2) lim_(x→−(1/2)) (1/(2x^3 +3x^2 +(√((1+2b+4bx)/4)))) × lim_(x→−(1/2)) (((2x^3 +3x^2 )^2 −(((1+2b+4bx)/4)))/((−2)(2x+1))) = −(3/4) −(1/8)×lim_(x→−(1/2)) ((4(2x^3 +3x^2 )^2 −1−2b−4bx)/(2x+1)) = −(3/4)
$$\left(\mathrm{1}\right)\:\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\sqrt{\mathrm{a}−\frac{\mathrm{b}}{\mathrm{2}}}\:=\:\mathrm{0} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\:=\:\sqrt{\mathrm{a}−\frac{\mathrm{b}}{\mathrm{2}}}\:\Rightarrow\:\mathrm{a}=\:\frac{\mathrm{1}+\mathrm{2b}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} +\sqrt{\frac{\mathrm{1}+\mathrm{2b}+\mathrm{4bx}}{\mathrm{4}}}}\:× \\ $$$$\underset{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\frac{\mathrm{1}+\mathrm{2b}+\mathrm{4bx}}{\mathrm{4}}\right)}{\left(−\mathrm{2}\right)\left(\mathrm{2x}+\mathrm{1}\right)}\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}}×\underset{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{4}\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{1}−\mathrm{2b}−\mathrm{4bx}}{\mathrm{2x}+\mathrm{1}}\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by abdomathmax last updated on 14/Mar/20
v(x)=4x^2 −1
$${v}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1} \\ $$
Commented by abdomathmax last updated on 14/Mar/20
⇒lim_(x→−(1/2)) ((2x^3 +3x^2 −(√(a+bx)))/(4x^2 −1)) +(3/4)=0 ⇒ lim_(x→−(1/2)) ((8x^3 +12x^2 −4(√(a+bx))+12x^2 −3)/(4x^2 −1))=0 ⇒ u(−(1/2))=0 and v(−(1/2))=0 with u(x)=8x^3 +24x^2 −4(√(a+bx)) −3 u(−(1/2))=0 ⇒8(−(1/8))+24((1/4))−4(√(a−(b/2)))−3=0 ⇒ −1+6−3 −4(√(a−(b/2)))=0 ⇒2−4(√(a−(b/2)))=0 ⇒ (√(a−(b/2)))=(1/2) ⇒a−(b/2) =(1/4) ⇒4a−2b = 1 ⇒ 4a+4b −6b =1 ⇒4(a+b) =1+6b ⇒a+b =((1+6b)/4)
$$\Rightarrow{lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\:\:\frac{\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{3}{x}^{\mathrm{2}} −\sqrt{{a}+{bx}}}{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}\:+\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\frac{\mathrm{8}{x}^{\mathrm{3}} \:+\mathrm{12}{x}^{\mathrm{2}} −\mathrm{4}\sqrt{{a}+{bx}}+\mathrm{12}{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{0}\:\Rightarrow \\ $$$${u}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}\:{and}\:{v}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}\:{with} \\ $$$${u}\left({x}\right)=\mathrm{8}{x}^{\mathrm{3}} \:+\mathrm{24}{x}^{\mathrm{2}} −\mathrm{4}\sqrt{{a}+{bx}}\:−\mathrm{3}\: \\ $$$${u}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}\:\Rightarrow\mathrm{8}\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)+\mathrm{24}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{4}\sqrt{{a}−\frac{{b}}{\mathrm{2}}}−\mathrm{3}=\mathrm{0}\:\Rightarrow \\ $$$$−\mathrm{1}+\mathrm{6}−\mathrm{3}\:−\mathrm{4}\sqrt{{a}−\frac{{b}}{\mathrm{2}}}=\mathrm{0}\:\Rightarrow\mathrm{2}−\mathrm{4}\sqrt{{a}−\frac{{b}}{\mathrm{2}}}=\mathrm{0}\:\Rightarrow \\ $$$$\sqrt{{a}−\frac{{b}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{a}−\frac{{b}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\mathrm{4}{a}−\mathrm{2}{b}\:=\:\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{4}{a}+\mathrm{4}{b}\:−\mathrm{6}{b}\:=\mathrm{1}\:\Rightarrow\mathrm{4}\left({a}+{b}\right)\:=\mathrm{1}+\mathrm{6}{b}\:\Rightarrow{a}+{b}\:=\frac{\mathrm{1}+\mathrm{6}{b}}{\mathrm{4}} \\ $$
Commented by jagoll last updated on 14/Mar/20

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