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Question Number 115103 by bobhans last updated on 23/Sep/20

lim_(x→∞) (√((x−a)(x+2))) −(√(x(x+1))) = 2 then a = ?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\left({x}−{a}\right)\left({x}+\mathrm{2}\right)}\:−\sqrt{{x}\left({x}+\mathrm{1}\right)}\:=\:\mathrm{2} \\ $$$${then}\:{a}\:=\:? \\ $$

Commented by Dwaipayan Shikari last updated on 23/Sep/20

lim_(x→∞) (((x−a)(x+2)−x^2 −x)/( x(√((1−(a/x))(1+(2/x))))+x(√(1+(1/x)))))=2 lim_(x→∞) ((x^2 −ax+2x−2a−x^2 −x)/(x+x))=2 lim_(x→∞) ((−a−1+2−((2a)/x))/2)=2 −a+1=4 a=−3 Or x(√((1−(a/x))(1+(2/x)))) −x(√(1+(1/x))) =2 x(1−(a/(2x))+(1/x)−(a/(2x^2 ))−1−(1/(2x)))=2 ((1−a)/2)−(a/(2x))=2 1−a=4 a=−3

$$\underset{{x}\rightarrow\infty} {{l}\mathrm{im}}\frac{\left({x}−{a}\right)\left({x}+\mathrm{2}\right)−{x}^{\mathrm{2}} −{x}}{\:{x}\sqrt{\left(\mathrm{1}−\frac{{a}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)}+{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}}=\mathrm{2} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −{ax}+\mathrm{2}{x}−\mathrm{2}{a}−{x}^{\mathrm{2}} −{x}}{{x}+{x}}=\mathrm{2} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{−{a}−\mathrm{1}+\mathrm{2}−\frac{\mathrm{2}{a}}{{x}}}{\mathrm{2}}=\mathrm{2} \\ $$$$−{a}+\mathrm{1}=\mathrm{4} \\ $$$${a}=−\mathrm{3} \\ $$$$ \\ $$$${Or} \\ $$$${x}\sqrt{\left(\mathrm{1}−\frac{{a}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)}\:−{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\:=\mathrm{2} \\ $$$${x}\left(\mathrm{1}−\frac{{a}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{{x}}−\frac{{a}}{\mathrm{2}{x}^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\right)=\mathrm{2} \\ $$$$\frac{\mathrm{1}−{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{2}{x}}=\mathrm{2} \\ $$$$\mathrm{1}−{a}=\mathrm{4} \\ $$$${a}=−\mathrm{3} \\ $$

Answered by bemath last updated on 23/Sep/20

lim_(x→∞) (√(x^2 +(2−a)x−2a))−(√(x^2 +x)) = 2 ⇒ ((2−a−1)/(2.1)) = 2 ; 1−a = 4 ⇒a = −3

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}^{\mathrm{2}} +\left(\mathrm{2}−{a}\right){x}−\mathrm{2}{a}}−\sqrt{{x}^{\mathrm{2}} +{x}}\:=\:\mathrm{2} \\ $$$$\Rightarrow\:\frac{\mathrm{2}−{a}−\mathrm{1}}{\mathrm{2}.\mathrm{1}}\:=\:\mathrm{2}\:;\:\mathrm{1}−{a}\:=\:\mathrm{4}\:\Rightarrow{a}\:=\:−\mathrm{3} \\ $$

Answered by ruwedkabeh last updated on 23/Sep/20

lim_(x→∞) (√((x−a)(x+2))) −(√(x(x+1))) = 2 ⇔lim_(x→∞) (√(x^2 +(2−a)x−2)) −(√(x^2 +x)) = 2 ((2−a−1)/(2(√1)))=2⇒((1−a)/2)=2⇒1−a=4⇒a=−3

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\left({x}−{a}\right)\left({x}+\mathrm{2}\right)}\:−\sqrt{{x}\left({x}+\mathrm{1}\right)}\:=\:\mathrm{2} \\ $$$$\:\Leftrightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}^{\mathrm{2}} +\left(\mathrm{2}−{a}\right){x}−\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} +{x}}\:=\:\mathrm{2} \\ $$$$\frac{\mathrm{2}−{a}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}}}=\mathrm{2}\Rightarrow\frac{\mathrm{1}−{a}}{\mathrm{2}}=\mathrm{2}\Rightarrow\mathrm{1}−{a}=\mathrm{4}\Rightarrow{a}=−\mathrm{3} \\ $$

Answered by Bird last updated on 24/Sep/20

let g(x) =(√((x−a)(x+2)))−(√(x(x+1))) ⇒g(x) =∣x∣(√(1−(a/x))).(√(1+(2/x))) −∣x∣(√(1+(1/x)))∼∣x∣{1−(a/(2x))} −∣x∣{1+(1/(2x))} =∣x∣−a ((∣x∣)/(2x))−∣x∣−((∣x∣)/(2x)) lim_(x→+∞) g(x)=2 ⇒ −(a/2)−(1/2) =2 ⇒−a−1 =4 ⇒ a+1 =−4 ⇒a=−5 lim_(x→−∞) g(x)=2 ⇒ (a/2)+(1/2)=2 ⇒a+1 =4 ⇒a=3

$${let}\:{g}\left({x}\right)\:=\sqrt{\left({x}−{a}\right)\left({x}+\mathrm{2}\right)}−\sqrt{{x}\left({x}+\mathrm{1}\right)} \\ $$$$\Rightarrow{g}\left({x}\right)\:=\mid{x}\mid\sqrt{\mathrm{1}−\frac{{a}}{{x}}}.\sqrt{\mathrm{1}+\frac{\mathrm{2}}{{x}}} \\ $$$$−\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\sim\mid{x}\mid\left\{\mathrm{1}−\frac{{a}}{\mathrm{2}{x}}\right\} \\ $$$$−\mid{x}\mid\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right\}\:=\mid{x}\mid−{a}\:\frac{\mid{x}\mid}{\mathrm{2}{x}}−\mid{x}\mid−\frac{\mid{x}\mid}{\mathrm{2}{x}} \\ $$$${lim}_{{x}\rightarrow+\infty} {g}\left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$$−\frac{{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{2}\:\Rightarrow−{a}−\mathrm{1}\:=\mathrm{4}\:\Rightarrow \\ $$$${a}+\mathrm{1}\:=−\mathrm{4}\:\Rightarrow{a}=−\mathrm{5} \\ $$$${lim}_{{x}\rightarrow−\infty} {g}\left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$$\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2}\:\Rightarrow{a}+\mathrm{1}\:=\mathrm{4}\:\Rightarrow{a}=\mathrm{3} \\ $$

Commented by bemath last updated on 24/Sep/20

if a = 3 lim_(x→∞) (√(x^2 −x−6))−(√(x^2 +x)) = ((−1−1)/(2.1))=−1 wrong. your answer wrong. if lim_(x→∞) ∣x∣ = lim_(x→∞) x , but lim_(x→−∞) ∣x∣ = lim_(x→−∞) −x

$${if}\:{a}\:=\:\mathrm{3}\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{6}}−\sqrt{{x}^{\mathrm{2}} +{x}}\:=\:\frac{−\mathrm{1}−\mathrm{1}}{\mathrm{2}.\mathrm{1}}=−\mathrm{1} \\ $$$${wrong}.\:{your}\:{answer}\:{wrong}. \\ $$$${if}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mid{x}\mid\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\:,\: \\ $$$${but}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\mid{x}\mid\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}−{x} \\ $$

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