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Question Number 149239 by john_santu last updated on 04/Aug/21

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

$$\:\:\:\underset{{x}\rightarrow\mathrm{a}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:=?\: \\ $$

Answered by EDWIN88 last updated on 04/Aug/21

 lim_(x→a^+ )  ((((√(x−a)))^2 )/(((√x)+(√a))(√((x−a)(x+a)))))−lim_(x→a^+ ) ((√(x−a))/( (√(x−a)) (√(x+a))))  = lim_(x→a^+ ) ((√(x−a))/(((√x)+(√a))(√(x+a))))−lim_(x→a^+ ) (1/( (√(x+a))))  = 0−(1/( (√(2a))))=−(1/( (√(2a))))

$$\:\underset{{x}\rightarrow{a}^{+} } {\mathrm{lim}}\:\frac{\left(\sqrt{{x}−{a}}\right)^{\mathrm{2}} }{\left(\sqrt{{x}}+\sqrt{{a}}\right)\sqrt{\left({x}−{a}\right)\left({x}+{a}\right)}}−\underset{{x}\rightarrow{a}^{+} } {\mathrm{lim}}\frac{\sqrt{{x}−{a}}}{\:\sqrt{{x}−{a}}\:\sqrt{{x}+{a}}} \\ $$$$=\:\underset{{x}\rightarrow{a}^{+} } {\mathrm{lim}}\frac{\sqrt{{x}−{a}}}{\left(\sqrt{{x}}+\sqrt{{a}}\right)\sqrt{{x}+{a}}}−\underset{{x}\rightarrow{a}^{+} } {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{{x}+{a}}} \\ $$$$=\:\mathrm{0}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}}}\: \\ $$

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