Question

Find the limit of the sequence using L'Hôpital's Rule.
\[d_n = n^{5}\left(\sqrt[6]{n^{6}+9}-n\right)\]

Answer 1

\[d_n = n^{5}\left(\sqrt[6]{n^{6}+9}-n\right)\]

Answer 2

cant i just guess? lol >.<

Answer 3

lol I wish. maybe im burning out tonight, but I dont even know how to manipulate it into the correct form..

Answer 4

yeah, me either....i know the answer is probably 0 though. i dont see how to apply L Hopital's Rule

Answer 5

it's not zero

Answer 6

It is not 0. fortunately I have unlimited submissions for this assignment.

Answer 7

\[\lim_{n \rightarrow \infty} (\sqrt[6]{n^{6}+9}-n)/(n ^{-5})\]
This is of the form 0/0
By L'Hopital's:
\[\lim_{n \rightarrow \infty} (n^{6}+9)^{-5/6}*6n^5/(-5n ^{-6})\]

Answer 8

\[\text{Limit}\left[n^5 \left(\sqrt[6]{n^6+9}-n\right),n\to \text{Infinity}\right]=\frac{3}{2} \]Does this help?

Answer 9

see told you it wasn't 0 ;)

Answer 10

well, robtobey you are correct but I dont know how you got that.

Answer 11

wah, i guess i have no guessing sense when it comes to these type of questions, drats >.< lol

Answer 12

unlike many, I dont come to this site to have others do my homework for me. I wont be here for the final.

Answer 13

do you see how to get it in the f(x)/g(x) form? if you do the rest is dull arithmeti

Answer 14

I "cheated". That is a Mathematica 8 Home Edition expression that I posted. WolframAlpha.com might show how they would solve it, step by step.

Answer 15

aha. nifty.

Answer 16

In ody's post above, he says its in the form 0/0. i think its actually the form -inf/0, but that is still valid right?

Answer 17

-inf over 0 is not valid for L'Hopitals (and sorry, in my post i forgot to include the -1)

Answer 18

-inf over 0 would be (in most cases) just -infinity

Answer 19

actully I think you are right, it is 0/0. I couldnt read the powers untill i switched to the web app i got this from.

Answer 20

simplify to this first

\[n^{5}\left(\sqrt[6]{n^{6}+9}-n\right)\]

\[n^{5}\left(\sqrt[6]{n^{6}\left(1+\frac{9}{n^6}\right)}-n\right)\]


\[n^{5}\left(n\sqrt[6]{1+\frac{9}{n^6}}-n\right)\]


\[n^{6}\left(\sqrt[6]{1+\frac{9}{n^6}}-1\right)\]

\[\frac{\left(\sqrt[6]{1+\frac{9}{n^6}}-1\right)}{n^{-6}}\]

use L'Hospitals rule on this

Answer 21

I could probably take it from there, but that is a lot of work and I already have an answer.