lim_(x→−∞) ((√(x^2 +3x+2))−x)


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Question Number 187859 by TUN last updated on 23/Feb/23

$\lim_{x \to - \infty} (\sqrt{x^{2} + 3 x + 2} - x)$
	Answered by Rasheed.Sindhi last updated on 23/Feb/23

	$\lim_{x \to - \infty} (\sqrt{x^{2} + 3 x + 2} - x)$ $\lim_{x \to - \infty} \frac{(\sqrt{x^{2} + 3 x + 2} - x) (\sqrt{x^{2} + 3 x + 2} + x)}{\sqrt{x^{2} + 3 x + 2} + x}$ $\lim_{x \to - \infty} \frac{x^{2} + 3 x + 2 - x^{2}}{\sqrt{x^{2} + 3 x + 2} + x}$ $\lim_{x \to - \infty} \frac{x (3 + \frac{2}{x})}{x (\sqrt{1 + \frac{3}{x} + \frac{2}{x^{2}}} + 1)}$ $\lim_{x \to - \infty} \frac{3 + \frac{2}{x}}{\sqrt{1 + \frac{3}{x} + \frac{2}{x^{2}}} + 1}$ $\frac{3 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{3}{2}$
	Commented by Frix last updated on 23/Feb/23

	$The mistake is for x < 0 :$ $\sqrt{1 + \frac{3}{x} + \frac{2}{x^{2}}} = \frac{\sqrt{x^{2} + 3 x + 2}}{∣ x ∣} = - \frac{\sqrt{x^{2} + 3 x + 2}}{x}$ $\Rightarrow \lim_{x \to - \infty} \frac{3 + \frac{2}{x}}{1 - \sqrt{1 + \frac{3}{x} + \frac{2}{x^{2}}}} = \frac{3 - 0}{1 - \sqrt{1 - 0 + 0}} undefined$
	Commented by Rasheed.Sindhi last updated on 23/Feb/23

	$T h a n k s s i r, I l e a r n t!$
	Answered by cortano12 last updated on 23/Feb/23

	$L = \lim_{x \to - \infty} \sqrt{x^{2} (1 + \frac{3}{x} + \frac{2}{x^{2}})} - x$ $= \lim_{x \to - \infty} - x [\sqrt{1 + \frac{3}{x} + \frac{2}{x^{2}}} + 1]$ $= \lim_{x \to 0} \frac{\sqrt{1 - 3 x - {2 x}^{2}} + 1}{x} = \infty$
	Answered by aba last updated on 23/Feb/23

	$\lim_{x \to - \infty} \sqrt{x^{2} + 3 x + 2} = + \infty \land \underset{x \to - \infty}{\lim x} = - \infty$ $\Rightarrow \lim_{x \to - \infty} (\sqrt{x^{2} + 3 x + 2} - x) = + \infty$
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