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lim-x-1-ax-a-b-3-x-1-3-2-find-a-and-b-without-L-hopital-rule-

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Question Number 82699 by jagoll last updated on 23/Feb/20
lim_(x→1) (((√(ax−a+b))−3)/(x−1)) = −(3/2) find a and b without L′hopital rule
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{{ax}−{a}+{b}}−\mathrm{3}}{{x}−\mathrm{1}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${find}\:{a}\:{and}\:{b}\:{without}\:{L}'{hopital}\:{rule} \\ $$
Commented by john santu last updated on 23/Feb/20
(1) limit must be (0/0) ⇒ (√b) = 3 , b = 9 (2) lim_(x→1) (((√(a(x−1)+9 ))−3)/(x−1)) = −(3/2) let x−1 = t ⇒ lim_(t→0) (((√(at+9))−3)/t) = −(3/2) lim_(t→0) ((at)/t) ×lim_(t→0) (1/( (√(at+9))+3)) = −(3/2) a × (1/6) = −(3/2) ⇒ a = −9
$$\left(\mathrm{1}\right)\:{limit}\:{must}\:{be}\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\Rightarrow\:\sqrt{{b}}\:=\:\mathrm{3}\:,\:{b}\:=\:\mathrm{9} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{{a}\left({x}−\mathrm{1}\right)+\mathrm{9}\:}−\mathrm{3}}{{x}−\mathrm{1}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${let}\:{x}−\mathrm{1}\:=\:{t}\:\Rightarrow\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{at}+\mathrm{9}}−\mathrm{3}}{{t}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{at}}{{t}}\:×\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{at}+\mathrm{9}}+\mathrm{3}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${a}\:×\:\frac{\mathrm{1}}{\mathrm{6}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{a}\:=\:−\mathrm{9}\: \\ $$
Commented by jagoll last updated on 23/Feb/20
thanks
$${thanks} \\ $$
Commented by mathmax by abdo last updated on 23/Feb/20
⇒lim_(x→1) (((√(ax−a+b))−3)/(x−1))+(3/2) =0 ⇒ lim_(x→1) ((2(√(ax−a+b))−6 +3x−3)/(2(x−1)))=0 ⇒ lim_(x→1) u(x)=0 and lim_(x→1) u^′ (x)=0 with u(x)=2(√(ax−a+b))+3x−9 lim_(x→1) u(x)=0 ⇒2(√b)−6 =0 ⇒(√b)=3 ⇒b=9 u^′ (x)=(a/( (√(ax−a+b)))) +3 lim_(x→1) u^′ (x)=0 ⇒(a/( (√b)))+3 =0 ⇒(a/3)+3 =0 ⇒(a/3)=−3 ⇒a=−9
$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}} \:\:\:\frac{\sqrt{{ax}−{a}+{b}}−\mathrm{3}}{{x}−\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{2}}\:=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \:\:\:\:\frac{\mathrm{2}\sqrt{{ax}−{a}+{b}}−\mathrm{6}\:+\mathrm{3}{x}−\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{1}\right)}=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} {u}\left({x}\right)=\mathrm{0}\:{and}\:{lim}_{{x}\rightarrow\mathrm{1}} {u}^{'} \left({x}\right)=\mathrm{0}\:{with} \\ $$$${u}\left({x}\right)=\mathrm{2}\sqrt{{ax}−{a}+{b}}+\mathrm{3}{x}−\mathrm{9} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} {u}\left({x}\right)=\mathrm{0}\:\Rightarrow\mathrm{2}\sqrt{{b}}−\mathrm{6}\:=\mathrm{0}\:\Rightarrow\sqrt{{b}}=\mathrm{3}\:\Rightarrow{b}=\mathrm{9} \\ $$$${u}^{'} \left({x}\right)=\frac{{a}}{\:\sqrt{{ax}−{a}+{b}}}\:+\mathrm{3} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \:{u}^{'} \left({x}\right)=\mathrm{0}\:\Rightarrow\frac{{a}}{\:\sqrt{{b}}}+\mathrm{3}\:=\mathrm{0}\:\Rightarrow\frac{{a}}{\mathrm{3}}+\mathrm{3}\:=\mathrm{0}\:\Rightarrow\frac{{a}}{\mathrm{3}}=−\mathrm{3}\:\Rightarrow{a}=−\mathrm{9} \\ $$

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