I assume #[x]# is the smallest integer bigger than #x#. In the following answer, we'll use the notation #ceil(x)#, called the ceiling function.
Let #f(x) = e^x/(ceil(x)+1)#. Since #x# is strictly bigger than #0#, this means that the domain of #f# is #(0,+oo)#.
As #x>0#, #ceil(x) > 1# and since #e^x# is always positive, #f# is always strictly bigger than #0# in its domain. It is important to note that #f# is not injective and is also not continuous at the natural numbers. To prove this, let #n# be a natural number:
#R_n=lim_(x->n^+) f(x) = lim_(x->n^+) e^x/(ceilx+1)#
Because #x>n#, #ceil(x) = n+1#.
#R_n= e^n/(n+2)#
#L_n=lim_(x->n^-) f(x) = lim_(x->n^-) e^x/(ceilx+1)#
Similarly, #ceil(x) = n#.
#L_n = e^n/(n+1)#
Since the left and right sided limits are not equal, #f# is not continuous at the integers. Also, #L>R# for all #n in NN#.
As #f# is increasing in intervals bounded by the positive integers, the "smallest values" per interval will be as #x# approaches the lower bound from the right.
Hence, the minimum value of #f# is going to be
#R_0=lim_(x->0^+) f(x) = lim_(x->0^+) e^x/(ceil(x)+1) = e^0/(0+2) = 1/2#
This is the lower bound of the range of #f#.
While it isn't truly correct to say that #f# is increasing, it is in the sense, asymptotically, it approaches infinity - as proved below:
#lim_(x->oo) f(x) =lim_(x->oo) e^x/(ceil(x)+1)#
As #ceilx >= x#, there exists a #delta<1# such that #ceilx=x+delta#:
#=lim_(x->oo) e^x/(x+delta+1)#
Let #u =x+ delta+1 => x=u-delta-1#.
#=lim_(u->oo) e^(u-delta-1)/u = [lim_(u->oo) e^u/u] * 1/e^(delta+1)#
#e^u# increases exponentially while #u# does so linearly, meaning that
#lim_(u->oo) e^u/u = oo#
#:. [lim_(u->oo) e^u/u] * 1/e^(delta+1) = oo * 1/e^(delta+1) = oo#
#:. lim_(x->oo) f(x) = oo#
Therefore the range of #f# is
#"Range" = (1/2, oo)#
The interval is open on the left because #1//2# is still #f(0)#, and as #x# approaches #0^+#, #f(x)# only approaches #1//2#; it never truly is equal.
