limx→∞ log[x]x where [x] denotes the greatest integer less than or equal to x, is
0
1
-1
non-existent
Let x=n+k, where 0<k<1. Then, [x]=n
∴ limx→∞ log[x]x=limn→∞ lognn+k⇒ limx→∞ log[x]x=limn→∞ 1n=0