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- anonymous

need help...first time on site:
limit to inifinity of this
n^.5 plus n^(1/3) divided by
n + 2n^(2/3)
book pulled out n to the neg .5
not sure why. Am thinking there is a better way?
Don't you always divide by highest power?

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- anonymous

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- ZeHanz

IMO, if you divide everything by n^(2/3), you get:\[\frac{ n^{-\frac{1}{6}} +n^{-\frac{1}{3}}}{ n^{\frac{1}{3}}+2 }\]Now, if n goes to infinity, the powers of n in the numerator go to 0, the denominator goes to infinity, so the whole thing goes to 0.
So, I think you're right!

- anonymous

|dw:1360016424558:dw|
I believe you divide by the lowest power, which is 1/3. so it would be.

- anonymous

|dw:1360016724321:dw|
At this point 1 and 2n^(1/3) do not matter.

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- anonymous

|dw:1360016863153:dw|
so the n's cancel out, that leaves (1/6)-(2/3) which equals -1/2

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