Consider the limit:
#(1) " " lim_(x->0) sin(1/x)#
If such limit exists, then for any sequence #{x_n}# such that:
#lim_(n->oo) x_n =0#
we should have:
#lim_(n->oo) sin(1/x_n) = lim_(x->0) sin(1/x)#
However if we choose #x_n = 1/(2pin)# then:
#lim_(n->oo) sin(1/x_n) = lim_(n->oo) sin(2pin) = 0#
while if we choose #x_n = 1/(2pin+pi/2)# then:
#lim_(n->oo) sin(1/x_n) = lim_(n->oo) sin(2pin+pi/2) = 1#
So the limit #(1)# does not exist and #f(x)# cannot be continuous for #x=0# whatever the value of #c# and not being continuous it is also not differentiable.
For #x!=0# on the other hand, #f(x)# is the composition of differentiable functions, so it is differentiable using the chain rule:
#d/dx sin(1/x) = cos(1/x) d/dx (1/x) = -1/x^2cos(1/x)#
