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I need to find $$\lim_{n\to\infty} \frac{n^{1296}}{6^n}$$ I have nothing yet, maybe sandwich theorem but I'm not sure. Thanks in advance for help.

A-Level Student
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shir
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  • See for example https://math.stackexchange.com/q/55468/42969. – Martin R Feb 07 '21 at 09:44
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    L'Hopitale rule 1296 times? – zkutch Feb 07 '21 at 09:47
  • The example was helpful tank you – shir Feb 07 '21 at 10:01
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    @ShirMoshe More like this : create the statement $\lim_{n \to \infty}\frac{n^k}{6^n} = 0$ for all integers $k$, to prove by mathematical induction. Prove the base case, and then use L'Hopital for the inductive case. This means you don't need to do L'Hopital 1296 times. – Sarvesh Ravichandran Iyer Feb 07 '21 at 10:16
  • FWIW, $n^{1296}=6^n$ at $n\approx 1.0013854 \text{ & } 6331.3374$ – PM 2Ring Feb 07 '21 at 10:24
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    @ShirMoshe If you are not willing to try what other suggests, what the point of posting the question? A-Level Student's answer solves the problem in that way. – jjagmath Feb 07 '21 at 12:03
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    Please share your thoughts on this limit problem here. Posting questions which are just problem statements is discouraged here. – Paramanand Singh Feb 07 '21 at 13:07
  • @TeresaLisbon, there is an alternative way, by taking the power of 1296 out common from the whole brakett........then apply l's hopital only once (as I cannot post it as an answer, hence this comment) – Aatmaj Feb 20 '21 at 16:01

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Let $$L=\lim_{n\to\infty}\frac{n^{1296}}{6^n}$$ Applying L' Hopital's rule once, we see that $$L=\lim_{n\to\infty}\frac{1296n^{1295}}{6^n\ln6}$$ We can keep on applying L' Hopital's rule (as we always will have a indeterminate form), until we get $$L=\lim_{n\to\infty}\frac{1296!~n}{6^n(\ln6)^{1295}}$$ Applying this once more we are left with $$L=\lim_{n\to\infty}\frac{1296!}{6^n(\ln6)^{1296}}=0$$ So we see that $L=0$. I hope that was helpful. If you have any questions please don't hesitate to ask :)

A-Level Student
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