anonymous
  • anonymous
Find the limit as x approaches infinity (y) and limit as x approaches negative infinity(y) y=sin(x)/(2(x)^(2)+x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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IrishBoy123
  • 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
Loser66
  • Loser66
wolfram said that both them =0 since sin is bounded and 2x^2+x is not.

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anonymous
  • anonymous
ok so now what?
anonymous
  • anonymous
do i plug in something for x or what? this is confusing
IrishBoy123
  • 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 \(\frac{1}{2x^2 + x } = \frac{ 1/x^2}{2 + 1/x } \)
anonymous
  • anonymous
wow limits are ridiculous
anonymous
  • anonymous
thnx @IrishBoy123

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