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