lim_(x→∞) (−1)^(x−1) sin (π(√(x^2 +0.5x+1))), where x∈N.


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Question Number 34439 by rahul 19 last updated on 06/May/18

$\lim_{x \to \infty} {(- 1)}^{x - 1} \sin (π \sqrt{x^{2} + 0.5 x + 1}),$ $w h e r e x \in N .$
	Answered by MJS last updated on 07/May/18
	$for great values of x: (√(x^2 +ax+b))=x+(a/2) x^2 +ax+b (x+(a/2))^2 =x^2 +ax+(a/4) lim_(x→∞) ((x^2 +ax+b)/(x^2 +ax+(a/4)))=1 sin π(√(x^2 +(1/2)x+1))=sin π(x+(1/4))= { ((−((√2)/2), x=2k+1, k∈N)),((((√2)/2), x=2k, k∈N)) :} lim_(x→∞) (−1)^(x−1) sin π(√(x^2 +(1/2)x+1))=−((√2)/2)$
	$for great values of x :$ $\sqrt{x^{2} + a x + b} = x + \frac{a}{2}$ $x^{2} + a x + b$ ${(x + \frac{a}{2})}^{2} = x^{2} + a x + \frac{a}{4}$ $\lim_{x \to \infty} \frac{x^{2} + a x + b}{x^{2} + a x + \frac{a}{4}} = 1$ $\sin π \sqrt{x^{2} + \frac{1}{2} x + 1} = \sin π (x + \frac{1}{4}) = {\begin{cases} - \frac{\sqrt{2}}{2}, x = 2 k + 1, k \in N \\ \frac{\sqrt{2}}{2}, x = 2 k, k \in N \end{cases}$ $\lim_{x \to \infty} {(- 1)}^{x - 1} \sin π \sqrt{x^{2} + \frac{1}{2} x + 1} = - \frac{\sqrt{2}}{2}$
	Commented by rahul 19 last updated on 07/May/18

	$T h e c o r r e c t a n s w e r i s - \frac{1}{\sqrt{2}}$
	Commented by MJS last updated on 07/May/18

	${you}^{'} re right, I^{'} ll correct my answer$ $- \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$
	Commented by rahul 19 last updated on 07/May/18
	as you have neglect b , can't we neglect " ax " also as it is much smaller than x² I mean in this case (x+1/4) , why you have not neglected 1/4 ?
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