The first by exhibiting a function valid on that interval, and the second by contradiction.
I assume you mean uniformly continuous
First note that 1/x - 1/y = y-x/xy
For the first part:
For x, y in B, we have x,y>1 so 1/xy < 1
Thus, for all x,y in B lf(x) - f(y)l = l1/x - 1/yl = lx-yl/lxyl < lx-yl
and uniform continuity is not hard to show (let delta = epsilon).
For the second part, you have to show that there exists an epsilon > 0, you can find an x, y small enough such that their difference is greater than epsilon.
To prove that 1/x is not uniformly continuous on (0, ) we must show that:
There exists ε>0 such that for all δ>0 there exists x,wε(0, ) such that:
|x-w|<δ and |f(x)-f(w)| .
Let ε=2.
Take 0<δ<1 .
Choose x= δ and w= δ/4 then we have:
|x-w| = |δ-
AND
|f(x)-f(w)| = .
Hence 1/x is not uniformly continuous on (0, )