lim_(x→∞) ((√(x^2 +8x+9)) −(√(x^2 +5x+4)) )^(4x) = ?


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Question Number 161867 by cortano last updated on 23/Dec/21

$\lim_{x \to \infty} {(\sqrt{x^{2} + 8 x + 9} - \sqrt{x^{2} + 5 x + 4})}^{4 x} = ?$
	Answered by Ar Brandon last updated on 23/Dec/21

	$A = \lim_{x \to \infty} {(\sqrt{x^{2} + 8 x + 9} - \sqrt{x^{2} + 5 x + 4})}^{4 x}$ $= \lim_{x \to \infty} {(x \sqrt{1 + \frac{8 x + 9}{x^{2}}} - x \sqrt{1 + \frac{5 x + 4}{x^{2}}})}^{4 x}$ $= \lim_{x \to \infty} {(x (1 + \frac{8 x + 9}{2 x^{2}}) - x (1 + \frac{5 x + 4}{2 x^{2}}))}^{4 x}$ $= \lim_{x \to \infty} {(x (\frac{3 x + 5}{2 x^{2}}))}^{4 x} = \lim_{x \to \infty} {(\frac{3 x + 5}{2 x s g n (x)})}^{4 x}$ $\ln A = \underset{x \to \infty}{\lim 4} x \ln (\frac{3 x + 5}{2 x}) = \underset{x \to \infty}{\lim 4} x [\ln (\frac{3}{2}) + \ln (1 + \frac{5}{3 x})]$ $= \underset{x \to \infty}{\lim 4} x (\ln (\frac{3}{2}) + \frac{5}{3 x}) = \underset{x \to \infty}{\lim 4} x \ln (\frac{3}{2}) \to \infty$ $\Rightarrow A \underset{x \to \infty}{\to} \infty$
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