lim_(x→∞) { (x/(x+(((x)^(1/3) )/(x+ (((x)^(1/3) )/(x+(((x)^(1/3) )/(......... infinity ))))))))}


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Question Number 34921 by rahul 19 last updated on 13/May/18
$lim_(x→∞) { (x/(x+(((x)^(1/3) )/(x+ (((x)^(1/3) )/(x+(((x)^(1/3) )/(......... infinity ))))))))}$
$\lim_{x \to \infty} {\frac{x}{x + \frac{{(x)}^{\frac{1}{3}}}{x + \frac{{(x)}^{\frac{1}{3}}}{x + \frac{{(x)}^{\frac{1}{3}}}{. . . . . . . . . i n f i n i t y}}}}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18
	$let D_r =t=x+(x^(1/3) /t) t^2 −tx−x^(1/3) =0 t=((x+(√(x^2 −4.1.(−x^(1/3) ))))/2) =((x+(√(x^2 +4x^(1/3) )))/2) lim_(x→∞ (/)) ((lim)/(x→∞)) ((x×2)/(x+(√(x^2 +4x^(1/3) )))) =((lim)/(x→∞ ))((x×2)/(x{1+(√(1+(4/x^(2−(1/3)) ))))) =((lim)/(x→∞ ))(2/({1+(√(1+(4/∞))))) =1$
	$l e t D_{r} = t = x + \frac{x^{\frac{1}{3}}}{t}$ $t^{2} - t x - x^{\frac{1}{3}} = 0$ $t = \frac{x + \sqrt{x^{2} - 4.1 . (- x^{\frac{1}{3}})}}{2}$ $= \frac{x + \sqrt{x^{2} + 4 x^{\frac{1}{3}}}}{2}$ $\underset{x \to \infty \frac{}{}}{l i m}$ $\frac{l i m}{x \to \infty} \frac{x \times 2}{x + \sqrt{x^{2} + 4 x^{\frac{1}{3}}}}$ $= \frac{l i m}{x \to \infty} \frac{x \times 2}{x {1 + \sqrt{1 + \frac{4}{x^{2 - \frac{1}{3}}}}}$ $= \frac{l i m}{x \to \infty} \frac{2}{{1 + \sqrt{1 + \frac{4}{\infty}}}$ $= 1$
	Commented by rahul 19 last updated on 13/May/18

	$T h a n k y o u s i r .$
	Commented by NECx last updated on 15/May/18

	$w o w$
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