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


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

$\lim_{x \to \infty} \sqrt{(x - a) (x + 2)} - \sqrt{x (x + 1)} = 2$ $t h e n a = ?$
Commented by Dwaipayan Shikari last updated on 23/Sep/20

$\underset{x \to \infty}{l im} \frac{(x - a) (x + 2) - x^{2} - x}{x \sqrt{(1 - \frac{a}{x}) (1 + \frac{2}{x})} + x \sqrt{1 + \frac{1}{x}}} = 2$ $\lim_{x \to \infty} \frac{x^{2} - a x + 2 x - 2 a - x^{2} - x}{x + x} = 2$ $\lim_{x \to \infty} \frac{- a - 1 + 2 - \frac{2 a}{x}}{2} = 2$ $- a + 1 = 4$ $a = - 3$ $O r$ $x \sqrt{(1 - \frac{a}{x}) (1 + \frac{2}{x})} - x \sqrt{1 + \frac{1}{x}} = 2$ $x (1 - \frac{a}{2 x} + \frac{1}{x} - \frac{a}{2 x^{2}} - 1 - \frac{1}{2 x}) = 2$ $\frac{1 - a}{2} - \frac{a}{2 x} = 2$ $1 - a = 4$ $a = - 3$
	Answered by bemath last updated on 23/Sep/20

	$\lim_{x \to \infty} \sqrt{x^{2} + (2 - a) x - 2 a} - \sqrt{x^{2} + x} = 2$ $\Rightarrow \frac{2 - a - 1}{2.1} = 2; 1 - a = 4 \Rightarrow a = - 3$
	Answered by ruwedkabeh last updated on 23/Sep/20

	$\lim_{x \to \infty} \sqrt{(x - a) (x + 2)} - \sqrt{x (x + 1)} = 2$ $\Leftrightarrow \lim_{x \to \infty} \sqrt{x^{2} + (2 - a) x - 2} - \sqrt{x^{2} + x} = 2$ $\frac{2 - a - 1}{2 \sqrt{1}} = 2 \Rightarrow \frac{1 - a}{2} = 2 \Rightarrow 1 - a = 4 \Rightarrow a = - 3$
	Answered by Bird last updated on 24/Sep/20
	$let g(x) =(√((x−a)(x+2)))−(√(x(x+1))) ⇒g(x) =∣x∣(√(1−(a/x))).(√(1+(2/x))) −∣x∣(√(1+(1/x)))∼∣x∣{1−(a/(2x))} −∣x∣{1+(1/(2x))} =∣x∣−a ((∣x∣)/(2x))−∣x∣−((∣x∣)/(2x)) lim_(x→+∞) g(x)=2 ⇒ −(a/2)−(1/2) =2 ⇒−a−1 =4 ⇒ a+1 =−4 ⇒a=−5 lim_(x→−∞) g(x)=2 ⇒ (a/2)+(1/2)=2 ⇒a+1 =4 ⇒a=3$
	$l e t g (x) = \sqrt{(x - a) (x + 2)} - \sqrt{x (x + 1)}$ $\Rightarrow g (x) =∣ x ∣ \sqrt{1 - \frac{a}{x}} . \sqrt{1 + \frac{2}{x}}$ $- ∣ x ∣ \sqrt{1 + \frac{1}{x}} \sim∣ x ∣ {1 - \frac{a}{2 x}}$ $- ∣ x ∣ {1 + \frac{1}{2 x}} =∣ x ∣ - a \frac{∣ x ∣}{2 x} - ∣ x ∣ - \frac{∣ x ∣}{2 x}$ ${l i m}_{x \to + \infty} g (x) = 2 \Rightarrow$ $- \frac{a}{2} - \frac{1}{2} = 2 \Rightarrow - a - 1 = 4 \Rightarrow$ $a + 1 = - 4 \Rightarrow a = - 5$ ${l i m}_{x \to - \infty} g (x) = 2 \Rightarrow$ $\frac{a}{2} + \frac{1}{2} = 2 \Rightarrow a + 1 = 4 \Rightarrow a = 3$
	Commented by bemath last updated on 24/Sep/20

	$i f a = 3$ $\lim_{x \to \infty} \sqrt{x^{2} - x - 6} - \sqrt{x^{2} + x} = \frac{- 1 - 1}{2.1} = - 1$ $w r o n g . y o u r a n s w e r w r o n g .$ $i f \lim_{x \to \infty} ∣ x ∣ = \lim_{x \to \infty} x,$ $b u t \lim_{x \to - \infty} ∣ x ∣ = \lim_{x \to - \infty} - x$
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