The function #f(x) = sin(e^x)# is defined and continuous in all of #RR#, so that for any #x_0#:
#lim_(x->x_0) f(x) = f(x_0)#
and in particular for #x_k = ln(kpi)# with #k in NN#:
#lim_(x->x_k) f(x) = f(x_k)#
which means that for every #k#, given #epsilon > 0# we can find #delta_(epsilon,k) > 0# such that:
#abs(x-x_k) < delta_(epsilon,k) => abs(f(x)-f(x_k)) < epsilon#
For the function to be evenly continuous, the choice of #delta_epsilon# must be independent of #x_k#.
Choose now #epsilon < 1# and suppose there is such a value of #delta_epsilon# that is good for every #x_k#, then we would have:
(1) #abs(x-x_k) < delta_(epsilon) => abs(f(x)-f(x_k)) < epsilon < 1# for any #k in NN#
Consider now:
#abs(f(x)-f(x_k)) = abs(sin(e^x)-sin(e^(x_k))#
as:
#sin(e^(x_k)) = sin(e^(ln(kpi) ))= sin(kpi) = 0#
we have:
#abs(f(x)-f(x_k)) = abs(sin(e^x))#
Take the interval #x in (x_k-delta_epsilon, x_k + delta_epsilon)# and express #x# as #x= x_k+xi#:
#abs(f(x)-f(x_k)) = abs(sin(e^(x_k+xi)))#
#abs(f(x)-f(x_k)) = abs(sin(e^(x_k)e^xi))#
#abs(f(x)-f(x_k)) = abs(sin(e^ln(pik)e^xi))#
#abs(f(x)-f(x_k)) = abs(sin(kpie^xi))#
Now pose: #u = kpie^xi#. As the exponential function is strictly increasing, as #xi# varies from #-delta_epsilon# to #delta_epsilon#, #u# varies from #kpie^(-delta_epsilon)# to #kpie^(delta_epsilon)#.
We can always choose #k# large enough that:
#pik(e^(delta_epsilon)-e^(-delta_epsilon)) > 2pi#
so that the interval covers more than one period of #sin u#, which means that there is at least one point in the interval where #sin u = 1#
In other words, we can always find a value of #k# such that there is a value of #x# for #x in (x_k-delta_epsilon, x_k+delta_epsilon)# such that:
#abs(f(x)-f(x_k)) = 1 > epsilon#
which is in contradiction with (1). We can conclude then that the value #delta_epsilon# depends on #k#, and then that the function is not uniformly continuous.
