Question

Let [x] be the floor function - the largest integer not exceeding x.

Answer 1

hint: {x} always lies between 0 and 1

Answer 2

thats what i dont get, why isit \[0\le{x}<1\]

Answer 3

hmm.. ok that makes it clear. i think i get it now if {x} lies between 0 and 1, then \[\frac{ 1 }{ \sqrt{x}}\] \[\ge0\] then its limit infinity is 0?

Answer 4

right.

Answer 5

okay thanks alot!(: