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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.