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

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

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(−\mathrm{1}\right)^{{x}−\mathrm{1}} \mathrm{sin}\:\left(\pi\sqrt{{x}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}{x}+\mathrm{1}}\right), \\ $$$${where}\:{x}\in\mathbb{N}. \\ $$

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)

$$\mathrm{for}\:\mathrm{great}\:\mathrm{values}\:\mathrm{of}\:{x}: \\ $$$$\sqrt{{x}^{\mathrm{2}} +{ax}+{b}}={x}+\frac{{a}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{ax}+{b} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{ax}+\frac{{a}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} +{ax}+{b}}{{x}^{\mathrm{2}} +{ax}+\frac{{a}}{\mathrm{4}}}=\mathrm{1} \\ $$$$\mathrm{sin}\:\pi\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}}=\mathrm{sin}\:\pi\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\begin{cases}{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\:{x}=\mathrm{2}{k}+\mathrm{1},\:{k}\in\mathbb{N}}\\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\:{x}=\mathrm{2}{k},\:{k}\in\mathbb{N}}\end{cases} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(−\mathrm{1}\right)^{{x}−\mathrm{1}} \mathrm{sin}\:\pi\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Commented by rahul 19 last updated on 07/May/18

The correct answer is −(1/(√2))

$${The}\:{correct}\:\:{answer}\:{is}\:−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$

Commented by MJS last updated on 07/May/18

you′re right, I′ll correct my answer −(1/(√2))=−((√2)/2)

$$\mathrm{you}'\mathrm{re}\:\mathrm{right},\:\mathrm{I}'\mathrm{ll}\:\mathrm{correct}\:\mathrm{my}\:\mathrm{answer} \\ $$$$−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

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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