lim_(x→∞) (√(4x^2 −16x+1))−2x+3=?


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Question Number 168004 by mathlove last updated on 31/Mar/22

$\lim_{x \to \infty} \sqrt{4 x^{2} - 16 x + 1} - 2 x + 3 = ?$
	Answered by nurtani last updated on 31/Mar/22

	$\lim_{x \to \infty} \sqrt{4 x^{2} - 16 x + 1} - 2 x + 3 = \lim_{x \to \infty} \sqrt{4 x^{2} - 16 x + 1} - (2 x - 3)$ $= \lim_{x \to \infty} \sqrt{4 x^{2} - 16 x + 1} - \sqrt{{(2 x - 3)}^{2}}$ $= \lim_{x \to \infty} \sqrt{4 x^{2} - 16 x + 1} - \sqrt{4 x^{2} - 12 x + 9}$ $= \lim_{x \to \infty} \sqrt{4 x^{2} - 16 x + 1} - \sqrt{4 x^{2} - 12 x + 9} \times \frac{\sqrt{4 x^{2} - 16 x + 1} + \sqrt{4 x^{2} - 12 x + 9}}{\sqrt{4 x^{2} - 16 x + 1 +} \sqrt{4 x^{2} - 12 x + 9}}$ $= \lim_{x \to \infty} \frac{4 x^{2} - 16 x + 1 - (4 x^{2} - 12 x + 9)}{\sqrt{4 x^{2} - 16 x + 1} + \sqrt{4 x^{2} - 12 x + 9}}$ $= \lim_{x \to \infty} \frac{4 x^{2} - 16 x + 1 - 4 x^{2} + 12 x - 9}{\sqrt{4 x^{2} - 16 x + 1} + \sqrt{4 x^{2} - 12 x + 9}}$ $= \lim_{x \to \infty} \frac{- 4 x - 8}{\sqrt{4 x^{2} - 16 x + 1} + \sqrt{4 x^{2} - 12 x + 9}}$ $= \frac{- 4}{\sqrt{4} + \sqrt{4}} = \frac{- 4}{2 + 2} = \frac{- 4}{4} = - 1$
	Commented by mathlove last updated on 31/Mar/22

	$t h a n k s$
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