Find $\lim_{n\to\infty} \frac{n^{1296}}{6^n}$

Question

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.

Answer 1

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 :)