limit as n goes to infinity of (ln(n)/(8^n))

Can someone explain how to find this?

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- Oct 26th 2012, 01:11 PMPreston019How to solve this limit
limit as n goes to infinity of (ln(n)/(8^n))

Can someone explain how to find this? - Oct 26th 2012, 01:40 PMskeeterRe: How to solve this limit
L'Hopital ...

- Oct 26th 2012, 02:00 PMjohnsomeoneRe: How to solve this limit
$\displaystyle \text{First, seeing } 8^n, \text{ you should reflexively see } 8^n = e^{\ln(8^n)} = e^{n \ln(8)} = e^{(\ln(8))n}$

$\displaystyle \text{Second, clearly for n greater than 1, it's positive. It should also be clear that it's decreasing, though that takes some work to prove.}$

$\displaystyle \text{Third, it should hopefully be obvious that the answer is 0, and that it's getting there very quickly.}$

$\displaystyle \text{There are 3 ways to the answer:}$

$\displaystyle \text{1) L'Hopital's Rule - assuming you know it. The answer drops right out.}$

$\displaystyle \text{2) Once you get familiar with these things, you'll know that "Logs lose to polynomials, hence certainly to exponentials."}$

$\displaystyle \text{In particular, if you can be assumed to know that }\lim_{n \to \infty} \frac{\ln(n)}{n} = 0,$

$\displaystyle \text{then it follows quickly after also observing that } n < 8^n \text{ for } n \ge 1.$

$\displaystyle \text{3) If neither of those approaches is available to you, you can solve it using the following idea:}$

$\displaystyle \text{Show that, for some fixed integer } N \text{ and for some fixed } r, \ 0 < r < 1,$

$\displaystyle \ a_n = \frac{\ln(n)}{8^n} \text{ satisfies: } 0 < a_{n+1} < r \ a_n \ \forall n \ge N .$

$\displaystyle \text{I would suggest choosing } r = 1/4. \text{ So determine when } a_{n+1} < \frac{1}{4}a_n.$

$\displaystyle \text{If you work out the algebra, it'll come down to showing that:}$

$\displaystyle n^2 - n - 1 >0 \text{ for } n \ge 2, \text{ which you should be able to do.}$