lim-x-1-x-1-sin-pi-x-2-0-5x-1-where-x-N-

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 :

\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 ?