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 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 + ...