Question

What is the limit of this function as x approaches negative infinite?

Answer 1

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

Answer 2

that's not the correct answer :(

Answer 3

hmmm... well. yes, I noticed that :|, the asymptote seems to be a bit lower than 0

Answer 4

are you supposed to graph it to get it?

Answer 5

can't use a graph on this one

Answer 6

If it helps they provided another version of this type of problem with the answer

Answer 7

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:
\[\lim_{x \rightarrow -\infty}\frac{ 3|x^{3}| }{ x^{3} }=\lim_{x \rightarrow -\infty}\frac{ 3|-\infty| }{ -\infty }=-3\]

Answer 8

Thank you so much :)

Answer 9

you're welcome. did that make any sense?