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Question-66518




Question Number 66518 by Masumsiddiqui399@gmail.com last updated on 16/Aug/19
Commented by mathmax by abdo last updated on 16/Aug/19
let f(x)=((x(√x)−a(√a))/(x−a))   cha7gement x−a=t give  lim_(x→a) f(x) =lim_(t→0)    (((t+a)(√(t+a))−a(√a))/t)  =lim_(t→0)    (((t+a)^3 −a^3 )/(t((t+a)(√(t+a)) +a(√a))))   =lim_(t→0)     ((t^3  +3t^2 a +3ta^2 )/(t{(t+a)(√(t+a))+a(√a)))) =lim_(t→0)  ((t^2  +3at +3a^2 )/((t+a)(√(t+a)) +a(√a)))  =((3a^2 )/(2a(√a))) =(3/2)(√a)
$${let}\:{f}\left({x}\right)=\frac{{x}\sqrt{{x}}−{a}\sqrt{{a}}}{{x}−{a}}\:\:\:{cha}\mathrm{7}{gement}\:{x}−{a}={t}\:{give} \\ $$$${lim}_{{x}\rightarrow{a}} {f}\left({x}\right)\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\left({t}+{a}\right)\sqrt{{t}+{a}}−{a}\sqrt{{a}}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\left({t}+{a}\right)^{\mathrm{3}} −{a}^{\mathrm{3}} }{{t}\left(\left({t}+{a}\right)\sqrt{{t}+{a}}\:+{a}\sqrt{{a}}\right)}\: \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\:\frac{{t}^{\mathrm{3}} \:+\mathrm{3}{t}^{\mathrm{2}} {a}\:+\mathrm{3}{ta}^{\mathrm{2}} }{{t}\left\{\left({t}+{a}\right)\sqrt{{t}+{a}}+{a}\sqrt{{a}}\right)}\:={lim}_{{t}\rightarrow\mathrm{0}} \:\frac{{t}^{\mathrm{2}} \:+\mathrm{3}{at}\:+\mathrm{3}{a}^{\mathrm{2}} }{\left({t}+{a}\right)\sqrt{{t}+{a}}\:+{a}\sqrt{{a}}} \\ $$$$=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{2}{a}\sqrt{{a}}}\:=\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{a}} \\ $$
Commented by Tony Lin last updated on 16/Aug/19
lim_(x→a) ((x(√x)−a(√a))/(x−a))  =lim_(x→a) ((x^(3/2) −a^(3/2) )/(x−a))   =lim_(x→a) (((3/2)x^(1/2) )/1)  =(3/2)(√a)
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}\sqrt{{x}}−{a}\sqrt{{a}}}{{x}−{a}} \\ $$$$=\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −{a}^{\frac{\mathrm{3}}{\mathrm{2}}} }{{x}−{a}}\: \\ $$$$=\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{a}} \\ $$

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