Saturday, February 11, 2017

limits - $limlimits_{nto infty}frac{n^k}{2^n}$ without l'Hopital




Let $\varepsilon >0$. Let $N>?$. For any integer $n>N$ we have
$$\frac{n^k}{2^n}<\varepsilon.$$
I don't know how to proceed here sensically.




I'd say we have to start with



$$<\frac{n^k}{2^N}$$



But what do we do here about the $n^k$?



Remember, I don't want to use l'Hopital or for me unproven limit laws like "exponents grow faster than powers"



Also, I don't want to use hidden l'Hopital, i.e. argumenting with derivatives. Since we don't even have proven derivative laws.


Answer




Let's take the logarithm. One gets



$$\log{n^k\over 2^n}=k\log{n}-n\log{2}=-n\left(\log{2}-k{\log{n}\over n}\right)$$



Now when $n\to\infty$ one has $\log{n}/n\to 0$ and so



$$\lim_{n\to\infty}\log{n^k\over 2^n}=-\infty$$



And so




$$\lim_{n\to\infty}{n^k\over 2^n}=0$$


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