lim_(x→∞) x((√(4x^2 −12x+7))−(√(x^2 −4x−2))−x+1) lim_(x→∞) ((√(x^2 +3x))−((x^3 +2x^2 ))^(1/3) )


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Question Number 186066 by TUN last updated on 31/Jan/23

$\lim_{x \to \infty} x (\sqrt{4 x^{2} - 12 x + 7} - \sqrt{x^{2} - 4 x - 2} - x + 1)$ $\lim_{x \to \infty} (\sqrt{x^{2} + 3 x} - \sqrt[3]{x^{3} + 2 x^{2}})$
	Answered by a.lgnaoui last updated on 31/Jan/23

	$x (\sqrt{4 x^{2} - 12 x + 7} - \sqrt{x^{2} - 4 x - 2} - x + 1)$ $= 2 x \sqrt{x^{2} - 3 x + \frac{7}{4}} - x \sqrt{x^{2} - 4 x - 2} - x^{2} + x$ $\lim_{x \to \infty} [2 x \sqrt{x^{2} - 3 x + \frac{7}{4}} - x \sqrt{x^{2} - 4 x - 2} - (x^{2} - x)]$ $= \lim_{x \to \infty} [x (\frac{2 (4 x - 3 x + \frac{15}{4})}{\sqrt{x^{2} - 3 x + \frac{7}{4}} + \sqrt{x^{2} - 4 x - 2}}) - x (x - 1)]$ $[(\frac{2 x^{2}}{x \sqrt{1 - \frac{3}{x} + \frac{7}{4 x^{2}}} + x \sqrt{1 - \frac{4}{x} - \frac{2}{x^{2}}}}) - x^{2} (1 - \frac{1}{x})]$ $= \lim_{x \to \infty} x - x^{2} (1 - \frac{1}{x}) = \lim_{x \to \infty} (- x^{2}) = - \infty$
	Commented by Frix last updated on 01/Feb/23

	$Wrong .$
	Answered by Frix last updated on 01/Feb/23

	$\lim_{x \to \infty} x (\sqrt{4 x^{2} - 12 x + 7} - \sqrt{x^{2} - 4 x - 2} - x + 1) =$ $\lim_{t \to 0^{+}} \frac{t - 1 + \sqrt{7 t^{2} - 12 t + 4} - \sqrt{- 2 t^{2} + 4 t + 1}}{t^{2}} =$ $= \lim_{t \to 0^{+}} \frac{\frac{d^{2} [t - 1 + \sqrt{7 t^{2} - 12 t + 4} - \sqrt{- 2 t^{2} + 4 t + 1}]}{{d t}^{2}}}{\frac{d^{2} [t^{2}]}{{d t}^{2}}} =$ $= \lim_{t \to 0^{+}} \frac{\frac{6}{{(- 2 t^{2} - 4 t + 1)}^{\frac{3}{2}}} - \frac{8}{{(7 t^{2} - 12 t + 4)}^{\frac{3}{2}}}}{2} =$ $= \frac{5}{2}$ $Similar procedure for the 2^{nd} one leads to \frac{5}{6}$
	Answered by cortano1 last updated on 01/Feb/23

	$W i t h o u t L^{'} H o p i t a l$ $L = \lim_{x \to \infty} x [\sqrt{4 x^{2} - 12 x + 7} - \sqrt{x^{2} - 4 x - 2} - x + 1]$ $= \lim_{x \to \infty} x [\sqrt{4 x^{2} - 12 x + 7} - (2 x - 3) + (x - 2) - \sqrt{x^{2} - 4 x - 2}]$ $= \lim_{x \to \infty} x [\frac{- 2}{\sqrt{4 x^{2} - 12 x + 7} + (2 x - 3)} + \frac{6}{x - 2 + \sqrt{x^{2} - 4 x - 2}}]$ $= \frac{- 2}{\sqrt{4} + 2} + \frac{6}{1 + 1} = \frac{5}{2}$
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