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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)
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)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439252879313:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\(\huge y=\frac{sin(x)}{2x^2+x}\) 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0wolfram said that both them =0 since sin is bounded and 2x^2+x is not.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do i plug in something for x or what? this is confusing

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i 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 \(\frac{1}{2x^2 + x } = \frac{ 1/x^2}{2 + 1/x } \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow limits are ridiculous
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