## anonymous one year ago Find the limit as x approaches infinity (y) and limit as x approaches negative infinity(y) y=sin(x)/(2(x)^(2)+x)

1. anonymous

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2. IrishBoy123

$$\huge y=\frac{sin(x)}{2x^2+x}$$ the numerator is always -1< ... < 1 the denominator looks like it would go nuts if $$x \rightarrow \pm \infty$$ and the $$x^2$$ term is dominant

3. Loser66

wolfram said that both them =0 since sin is bounded and 2x^2+x is not.

4. anonymous

ok so now what?

5. anonymous

do i plug in something for x or what? this is confusing

6. IrishBoy123

i think you need some maths b.s to finish it off you could say that $$lim_{x\rightarrow \pm \infty}$$ $$\frac{-1}{2x^2 + x } \le \frac{sin (x)}{2x^2 + x} \le \frac{1}{2x^2 + x}$$ $$lim_{x\rightarrow \pm \infty}$$ $$\frac{1}{2x^2 + x } = 0$$ $$\therefore lim_{x\rightarrow \pm \infty}$$ $$\frac{sin (x)}{2x^2 + x} = 0$$ a squeeze something like that. bearing in mind that $$\frac{1}{2x^2 + x } = \frac{ 1/x^2}{2 + 1/x }$$

7. anonymous

wow limits are ridiculous

8. anonymous

thnx @IrishBoy123