lim_(x→2) ((((5x−9))^(1/5) ((4x−7))^(1/4) ((3x−5))^(1/3) (√(2x−3))−1)/(x−2))=?


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Question Number 175137 by cortano1 last updated on 20/Aug/22

$\lim_{x \to 2} \frac{\sqrt[5]{5 x - 9} \sqrt[4]{4 x - 7} \sqrt[3]{3 x - 5} \sqrt{2 x - 3} - 1}{x - 2} = ?$
	Answered by Ar Brandon last updated on 20/Aug/22

	$L = \lim_{x \to 2} \frac{\sqrt[5]{5 x - 9} \sqrt[4]{4 x - 7} \sqrt[3]{3 x - 5} \sqrt{2 x - 3} - 1}{x - 2}$ $= \lim_{t \to 0} \frac{(\sqrt[5]{5 t + 1}) (\sqrt[4]{4 t + 1}) (\sqrt[3]{3 t + 1}) (\sqrt{2 t + 1}) - 1}{t}$ $= \lim_{t \to 0} \frac{1}{t} [(1 + t) (1 + t) (1 + t) (1 + t) - 1]$ $= \lim_{t \to 0} \frac{1}{t} [{(t^{2} + 2 t + 1)}^{2} - 1]$ $= \lim_{t \to 0} \frac{1}{t} (1 + 2 (t^{2} + 2 t) - 1)$ $= \lim_{t \to 0} \frac{1}{t} (2 t^{2} + 4 t) = 4$
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