lim_(x→∞) ((((x+1))^(1/4) − (x)^(1/4) )/( ((x+1))^(1/3) − (x)^(1/3) )) =?


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Question Number 124802 by john_santu last updated on 06/Dec/20

$\lim_{x \to \infty} \frac{\sqrt[4]{x + 1} - \sqrt[4]{x}}{\sqrt[3]{x + 1} - \sqrt[3]{x}} = ?$
	Answered by bramlexs22 last updated on 06/Dec/20

	$\lim_{x \to \infty} \frac{\sqrt[4]{x} (\sqrt[4]{1 + \frac{1}{x}} - 1)}{\sqrt[3]{x} (\sqrt[3]{1 + \frac{1}{x}} - 1)} =$ $\lim_{x \to \infty} \frac{1}{\sqrt[12]{x}} (\frac{\sqrt[4]{1 + \frac{1}{x}} - 1}{\sqrt[3]{1 + \frac{1}{x}} - 1}) = 0$
	Answered by Bird last updated on 06/Dec/20
	$f(x)=(((x+1)^(1/4) −x^(1/4) )/((x+1)^(1/3) −x^(1/3) )) ⇒ f(x)=((x^(1/4) {(1+(1/x))^(1/4) −1})/(x^(1/3) {(1+(1/x))^(1/3) −1})) ∼(1/x^((1/3)−(1/4)) )×((1/(4x))/(1/(3x))) (x→+∞) f(x)∼(3/4)×(1/x^(1/(12)) )→0 (x→+∞) ⇒lim_(x→+∞) f(x)=0$
	$f (x) = \frac{{(x + 1)}^{\frac{1}{4}} - x^{\frac{1}{4}}}{{(x + 1)}^{\frac{1}{3}} - x^{\frac{1}{3}}} \Rightarrow$ $f (x) = \frac{x^{\frac{1}{4}} {{(1 + \frac{1}{x})}^{\frac{1}{4}} - 1}}{x^{\frac{1}{3}} {{(1 + \frac{1}{x})}^{\frac{1}{3}} - 1}}$ $\sim \frac{1}{x^{\frac{1}{3} - \frac{1}{4}}} \times \frac{\frac{1}{4 x}}{\frac{1}{3 x}} (x \to + \infty)$ $f (x) \sim \frac{3}{4} \times \frac{1}{x^{\frac{1}{12}}} \to 0 (x \to + \infty)$ $\Rightarrow {l i m}_{x \to + \infty} f (x) = 0$
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