Can you do this without the use of Lhopital's rule

Similar approach using the Binomial Formula:

2^n = (1 + 1)^n >= n * (n+1) / 2 for n in N

With this approximation you get for x in [ 2^n; 2^{n+1) ):

0 <= log(x) / x <= log( 2^{n+1} ) / (1 + 1)^n <= 2 * log(2) / n -> 0

Via sandwich-lemma, your limit is zero.

Edit: Is there LaTeX support in this forum? Manual markdown feels kind of dirty...

set x = e^t and take limit t goes to infinity.

e^t = 1 + t + t^2/2 + ...

Similar approach using the Binomial Formula:

2^n = (1 + 1)^n >= n * (n+1) / 2 for n in NWith this approximation you get for

x in [ 2^n; 2^{n+1) ):0 <= log(x) / x <= log( 2^{n+1} ) / (1 + 1)^n <= 2 * log(2) / n -> 0Via sandwich-lemma, your limit is zero.

Edit: Is there LaTeX support in this forum? Manual markdown feels kind of dirty...

set x = e^t and take limit t goes to infinity.

e^t = 1 + t + t^2/2 + ...