anonymous
  • anonymous
What is the limit of this function as x approaches negative infinite?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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jdoe0001
  • jdoe0001
well, one can say that in the denominator, the if you give \(\bf x^3\) a negative value, it will always spit out a negative root, because the radical is ODD the bigger the negative value, the bigger the negative root on the numerator, \(\bf x^6\) is an EVEN exponent, and no matter what value you throw at it, it will always give a positive root so one can say the numerator will end up yielding a smaller number than the denominator so it will be a fraction, and due to the \(\bf \pm\) of the root, you'll have 2 fractions from each anyhow, after all that ramble, we can just look at the "degree" of the denominator and see is greater than that of the numerator when such happens, the "horizontal asymptote" occurs at y=0 or the x-axis so I'd think the fraction is approaching the x-axis or y = 0, or approaching 0
anonymous
  • anonymous
that's not the correct answer :(

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jdoe0001
  • jdoe0001
hmmm... well. yes, I noticed that :|, the asymptote seems to be a bit lower than 0
jdoe0001
  • jdoe0001
are you supposed to graph it to get it?
anonymous
  • anonymous
can't use a graph on this one
anonymous
  • anonymous
If it helps they provided another version of this type of problem with the answer
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anonymous
  • anonymous
it should go to -3... in the numerator, the 9x^6 term dominates but it's under the square root. so being to an even power will make it positive but taking the root will make it's power drop to 3. the sqrt of 9 is of course 3. so what happens is we look at the limit of this: \[\lim_{x \rightarrow -\infty}\frac{ 3|x^{3}| }{ x^{3} }=\lim_{x \rightarrow -\infty}\frac{ 3|-\infty| }{ -\infty }=-3\]
anonymous
  • anonymous
Thank you so much :)
anonymous
  • anonymous
you're welcome. did that make any sense?

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