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Question Number 28480 by naka3546 last updated on 26/Jan/18

Commented by math solver last updated on 29/Jan/18

$w h a t i s f u l l f o r m o f s b m p t n ?$
	Answered by ajfour last updated on 26/Jan/18
	$lim_(x→∞) [(2^(3x) +3^x )^(1/3) −(2^(2x) −2^x )^(1/2) ] =lim_(x→∞) 2^x {[1+((3/8))^x ]^(1/3) −[1−((1/2))^x ]^(1/2) } =lim_(x→∞) 2^x {1+(1/3)((3/8))^x +...−1+(1/2)((1/2))^x +...} =lim_(x→∞) [((3/4))^x +(1/2)] =(1/2) ⇒ (D).$
	$\lim_{x \to \infty} [{(2^{3 x} + 3^{x})}^{1 / 3} - {(2^{2 x} - 2^{x})}^{1 / 2}]$ $= {\underset{x \to \infty}{\lim 2}}^{x} {{[1 + {(\frac{3}{8})}^{x}]}^{1 / 3} - {[1 - {(\frac{1}{2})}^{x}]}^{1 / 2}}$ $= {\underset{x \to \infty}{\lim 2}}^{x} {1 + \frac{1}{3} {(\frac{3}{8})}^{x} + . . . - 1 + \frac{1}{2} {(\frac{1}{2})}^{x} + . . .}$ $= \lim_{x \to \infty} [{(\frac{3}{4})}^{x} + \frac{1}{2}] = \frac{1}{2} \Rightarrow (D) .$
	Commented by naka3546 last updated on 02/Feb/18
	$Is it using Taylor Series ? =lim_(x→∞) 2^x {[1+((3/8))^x ]^(1/3) −[1−((1/2))^x ]^(1/2) }$
	$I s i t u s i n g T a y l o r S e r i e s ?$ $= {\underset{x \to \infty}{\lim 2}}^{x} {{[1 + {(\frac{3}{8})}^{x}]}^{1 / 3} - {[1 - {(\frac{1}{2})}^{x}]}^{1 / 2}}$
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