anonymous
  • 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?
Calculus1
chestercat
  • chestercat
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ZeHanz
  • 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
  • anonymous
|dw:1360016424558:dw| I believe you divide by the lowest power, which is 1/3. so it would be.
anonymous
  • anonymous
|dw:1360016724321:dw| At this point 1 and 2n^(1/3) do not matter.

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

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