lim_(x→∞) ((27^x +9^x ))^(1/3) −(√(9^x +3^x )) =?


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Question Number 153352 by liberty last updated on 06/Sep/21

$\lim_{x \to \infty} \sqrt[3]{27^{x} + 9^{x}} - \sqrt{9^{x} + 3^{x}} = ?$
	Answered by MJS_new last updated on 06/Sep/21

	$\lim_{x \to + \infty} {(27^{x} + 9^{x})}^{1 / 3} - {(9^{x} + 3^{x})}^{1 / 2} =$ $[x = - \frac{\ln t}{\ln 3} \to t \to 0^{+}]$ $= \lim_{t \to 0^{+}} \frac{{(t + 1)}^{1 / 3} - {(t + 1)}^{1 / 2}}{t} =$ $[L^{'} H \hat{o p i t a l}]$ $= \lim_{t \to 0^{+}} \frac{2 - 3 {(t + 1)}^{1 / 6}}{6 {(t + 1)}^{2 / 3}} = - \frac{1}{6}$
	Commented by liberty last updated on 06/Sep/21

	$y e s s i r$
	Commented by MJS_new last updated on 06/Sep/21
	��
	Commented by liberty last updated on 06/Sep/21

	$h o w a b o u t f o r t h i s$ $\lim_{x \to \infty} \sqrt[3]{8^{x} + 3^{x}} - \sqrt{4^{x} - 2^{x}} = ?$ $r e s u l t 0 o r - \infty ?$
	Commented by MJS_new last updated on 07/Sep/21

	$it seems to be \frac{1}{2}$
	Commented by liberty last updated on 07/Sep/21

	$y e s s i r$
	Answered by liberty last updated on 06/Sep/21

	$\lim_{x \to \infty} \sqrt[3]{9^{x} (3^{x} + 1)} - \sqrt{3^{x} (3^{x} + 1)}$ $l e t 3^{x} + 1 = u^{6} \Rightarrow 3^{x} = u^{6} - 1$ $\lim_{u \to \infty} \sqrt[3]{{(3^{x})}^{2}} u^{2} - \sqrt{3^{x}} u^{3} =$ $\lim_{u \to \infty} u^{2} \sqrt[3]{{(u^{6} - 1)}^{2}} - u^{3} \sqrt{u^{6} - 1} =$ $\lim_{u \to \infty} u^{6} \sqrt[3]{{(1 - \frac{1}{u^{6}})}^{2}} - u^{6} \sqrt{1 - \frac{1}{u^{6}}} =$ $l e t \frac{1}{u^{6}} = t \land t \to 0$ $\lim_{t \to 0} \frac{\sqrt[3]{{(1 - t)}^{2}} - \sqrt{1 - t}}{t} =$ $\lim_{t \to 0} - \frac{2}{3} {(1 - t)}^{- \frac{1}{3}} + \frac{1}{2 \sqrt{1 - t}} =$ $- \frac{2}{3} + \frac{1}{2} = - \frac{4}{6} + \frac{3}{6} = - \frac{1}{6}$
	Answered by Aras last updated on 06/Sep/21


	Answered by bramlexs22 last updated on 07/Sep/21

	${\underset{x \to \infty}{\lim 3}}^{x} (\sqrt[3]{1 + \frac{1}{3^{x}}} - \sqrt{1 + \frac{1}{3^{x}}})$ $= \lim_{x \to 0} \frac{\sqrt[3]{+ x} - \sqrt{1 + x}}{x} = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}$
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