For any #x > 0# we have #absx/x = 1# while for any #x < 0# we have #absx/x = -1#.
We therefore have:
#lim_(x->0^+) absx/x = 1#
#lim_(x->0^-) absx/x = -1#
and as the limits frm the right and from the left do not coincide, the limit cannot exist.
To see it in terms of #epsilon-delta# we can use Cauchy' necessary condition, stating that:
#lim_(x->0) f(x)#
exists and is finite only if given a number #epsilon# we can find a #delta_epsilon# such that:
#(1) " " abs(f(x_1)-f(x_2)) < epsilon#
for any #x_1,x_2 in (-delta_epsilon, delta_epsilon)#.
However, if we take #epsilon = 1#, then for any #delta# we can choose #x_1 in (0,delta)# and #x_2 in (-delta,0)#, so that both points belong to #(-delta, delta)# but:
#abs ( abs(x_1)/x_1 -abs(x_2)/x_2) = abs( 1-(-1)) = 2 > 1#
So that no #delta# can be found to satisfy condition #(1)#.
On the other hand as the function is bounded the limit cannot be #+-oo# either.
So we can conclude that the limit does not exists.