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

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

Commented byDwaipayan 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

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

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

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

Commented bybemath 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