Question Number 84083 by john santu last updated on 09/Mar/20
$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}\:}\right)}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:= \\ $$
Commented by mathmax by abdo last updated on 09/Mar/20
$${let}\:{f}\left({a}\right)=\frac{\sqrt{{x}}−\sqrt{{a}}−\sqrt{{x}−{a}}}{\:\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:\:{changement}\:\sqrt{{x}−{a}}={t}\:{give} \\ $$$${x}−{a}={t}^{\mathrm{2}} \:\Rightarrow{a}={x}−{t}^{\mathrm{2}} \:\:\:\:\:{a}\rightarrow{x}\:\Leftrightarrow{t}\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${f}\left({a}\right)\:={g}\left({t}\right)=\frac{\sqrt{{x}}−\sqrt{{x}−{t}^{\mathrm{2}} }−{t}}{{t}\sqrt{{x}+{x}−{t}^{\mathrm{2}} }}\:=\frac{\sqrt{{x}}−\sqrt{{x}−{t}^{\mathrm{2}} }−{t}}{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }} \\ $$$$=\frac{\sqrt{{x}}−{t}−\sqrt{{x}−{t}^{\mathrm{2}} }}{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }}\:=\frac{\left(\sqrt{{x}}−{t}−\sqrt{{x}−{t}^{\mathrm{2}} }\right)\left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)}{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }\left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\left(\sqrt{{x}}−{t}\right)^{\mathrm{2}} −{x}+{t}^{\mathrm{2}} }{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} \left(\sqrt{{x}}−{t}+\sqrt{\left.{x}−{t}^{\mathrm{2}} \right)}\right.}}\:=\frac{{x}−\mathrm{2}{t}\sqrt{{x}}+{t}^{\mathrm{2}} −{x}\:+{t}^{\mathrm{2}} }{{t}\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} }\left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)} \\ $$$$=\frac{−\mathrm{2}\sqrt{{x}}−\mathrm{2}{t}}{\:\sqrt{\mathrm{2}{x}−{t}^{\mathrm{2}} \left(\sqrt{{x}}−{t}+\sqrt{{x}−{t}^{\mathrm{2}} }\right)}}\:\rightarrow\frac{−\mathrm{2}\sqrt{{x}}}{\:\sqrt{\mathrm{2}{x}}\left(\mathrm{2}\sqrt{{x}}\right)}\:=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}{x}}}\:\left({t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{a}\rightarrow{x}} \:{f}\left({x}\right)\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}}} \\ $$
Commented by john santu last updated on 10/Mar/20
$${thank}\:{you} \\ $$
Commented by mathmax by abdo last updated on 11/Mar/20
$${you}\:{are}\:{welcome} \\ $$
Answered by jagoll last updated on 09/Mar/20
$$\mathrm{i}\:\mathrm{use}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{a}}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{a}}}}{\frac{−\mathrm{2a}}{\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}}\:= \\ $$$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{−\mathrm{a}}\:×\:\frac{\mathrm{2}\sqrt{\mathrm{a}}\:−\mathrm{2}\sqrt{\mathrm{x}−\mathrm{a}}}{\mathrm{4}\sqrt{\mathrm{a}}\:\sqrt{\mathrm{x}−\mathrm{a}}}\:=\: \\ $$$$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}−\mathrm{a}\:}\:\sqrt{\mathrm{x}+\mathrm{a}}\:}{−\mathrm{a}}×\:\frac{\mathrm{2}\left(\sqrt{\mathrm{a}\:}−\sqrt{\mathrm{x}−\mathrm{a}\:}\:\right)}{\mathrm{4}\sqrt{\mathrm{a}\:}\:\sqrt{\mathrm{x}−\mathrm{a}}} \\ $$$$=\:\frac{\sqrt{\mathrm{2x}}}{−\mathrm{x}}\:×\:\frac{\mathrm{2}\sqrt{\mathrm{x}}}{\mathrm{4}\sqrt{\mathrm{x}}}\:=\:−\frac{\sqrt{\mathrm{2x}}}{\mathrm{2x}} \\ $$
Commented by john santu last updated on 10/Mar/20
$${good} \\ $$