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Question Number 84083 by john santu last updated on 09/Mar/20

lim_(a→x) ((((√x) −(√a) −(√(x−a ))))/(√(x^2 −a^2 ))) =

$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}\:}\right)}{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:= \\ $$

Commented by mathmax by abdo last updated on 09/Mar/20

let f(a)=(((√x)−(√a)−(√(x−a)))/(√(x^2 −a^2 ))) changement (√(x−a))=t give x−a=t^2 ⇒a=x−t^2 a→x ⇔t→0 ⇒ f(a) =g(t)=(((√x)−(√(x−t^2 ))−t)/(t(√(x+x−t^2 )))) =(((√x)−(√(x−t^2 ))−t)/(t(√(2x−t^2 )))) =(((√x)−t−(√(x−t^2 )))/(t(√(2x−t^2 )))) =((((√x)−t−(√(x−t^2 )))((√x)−t+(√(x−t^2 ))))/(t(√(2x−t^2 ))((√x)−t+(√(x−t^2 ))))) =((((√x)−t)^2 −x+t^2 )/(t(√(2x−t^2 ((√x)−t+(√(x−t^2 ))))))) =((x−2t(√x)+t^2 −x +t^2 )/(t(√(2x−t^2 ))((√x)−t+(√(x−t^2 ))))) =((−2(√x)−2t)/(√(2x−t^2 ((√x)−t+(√(x−t^2 )))))) →((−2(√x))/((√(2x))(2(√x)))) =((−1)/(√(2x))) (t→0) ⇒ lim_(a→x) f(x) =−(1/(√(2x)))

$${let}\:{f}\left({a}\right)=\frac{\sqrt{{x}}−\sqrt{{a}}−\sqrt{{x}−{a}}}{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:\:{changement}\:\sqrt{{x}−{a}}={t}\:{give} \\ $$$${x}−{a}={t}^{\mathrm{2}} \:\Rightarrow{a}={x}−{t}^{\mathrm{2}} \:\:\:\:\:{a}\rightarrow{x}\:\Leftrightarrow{t}\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${f}\left({a}\right)\:={g}\left({t}\right)=\frac{\sqrt{{x}}−\sqrt{{x}−{t}^{\mathrm{2}} }−{t}}{{t}\sqrt{{x}+{x}−{t}^{\mathrm{2}} }}\:=\frac{\sqrt{{x}}−\sqrt{{x}−{t}^{\mathrm{2}} }−{t}}{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }} \\ $$$$=\frac{\sqrt{{x}}−{t}−\sqrt{{x}−{t}^{\mathrm{2}} }}{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }}\:=\frac{\left(\sqrt{{x}}−{t}−\sqrt{{x}−{t}^{\mathrm{2}} }\right)\left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)}{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }\left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\left(\sqrt{{x}}−{t}\right)^{\mathrm{2}} −{x}+{t}^{\mathrm{2}} }{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} \left(\sqrt{{x}}−{t}+\sqrt{\left.{x}−{t}^{\mathrm{2}} \right)}\right.}}\:=\frac{{x}−\mathrm{2}{t}\sqrt{{x}}+{t}^{\mathrm{2}} −{x}\:+{t}^{\mathrm{2}} }{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }\left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)} \\ $$$$=\frac{−\mathrm{2}\sqrt{{x}}−\mathrm{2}{t}}{\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} \left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)}}\:\rightarrow\frac{−\mathrm{2}\sqrt{{x}}}{\sqrt{\mathrm{2}{x}}\left(\mathrm{2}\sqrt{{x}}\right)}\:=\frac{−\mathrm{1}}{\sqrt{\mathrm{2}{x}}}\:\left({t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{a}\rightarrow{x}} \:{f}\left({x}\right)\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}}} \\ $$

Commented by john santu last updated on 10/Mar/20

thank you

$${thank}\:{you} \\ $$

Commented by mathmax by abdo last updated on 11/Mar/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Answered by jagoll last updated on 09/Mar/20

i use L′Hopital lim_(a→x) ((−(1/(2(√a)))+(1/(2(√(x−a)))))/((−2a)/(2(√(x^2 −a^2 ))))) = lim_(a→x) ((√(x^2 −a^2 ))/(−a)) × ((2(√a) −2(√(x−a)))/(4(√a) (√(x−a)))) = lim_(a→x) (((√(x−a )) (√(x+a)) )/(−a))× ((2((√(a ))−(√(x−a )) ))/(4(√(a )) (√(x−a)))) = ((√(2x))/(−x)) × ((2(√x))/(4(√x))) = −((√(2x))/(2x))

$$\mathrm{i}\:\mathrm{use}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{a}}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{a}}}}{\frac{−\mathrm{2a}}{\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}}\:= \\ $$$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{−\mathrm{a}}\:×\:\frac{\mathrm{2}\sqrt{\mathrm{a}}\:−\mathrm{2}\sqrt{\mathrm{x}−\mathrm{a}}}{\mathrm{4}\sqrt{\mathrm{a}}\:\sqrt{\mathrm{x}−\mathrm{a}}}\:=\: \\ $$$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}−\mathrm{a}\:}\:\sqrt{\mathrm{x}+\mathrm{a}}\:}{−\mathrm{a}}×\:\frac{\mathrm{2}\left(\sqrt{\mathrm{a}\:}−\sqrt{\mathrm{x}−\mathrm{a}\:}\:\right)}{\mathrm{4}\sqrt{\mathrm{a}\:}\:\sqrt{\mathrm{x}−\mathrm{a}}} \\ $$$$=\:\frac{\sqrt{\mathrm{2x}}}{−\mathrm{x}}\:×\:\frac{\mathrm{2}\sqrt{\mathrm{x}}}{\mathrm{4}\sqrt{\mathrm{x}}}\:=\:−\frac{\sqrt{\mathrm{2x}}}{\mathrm{2x}} \\ $$

Commented by john santu last updated on 10/Mar/20

good

$${good} \\ $$

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