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.
