#f(x) = (x^3-3) # is a polynomial function therefore continuous in all of #RR#, so we know that:
#lim_(x->2) f(x) = f(2) = 5#
To give a formal proof we analyze the expression #abs (f(x) -5)# around #x=2# expressing #x# as #2+Deltax#:
#abs (f(x) -5) = abs ((2+Deltax)^3-3-5) =abs (8+12(Deltax)+6(Deltax)^2+(Deltax)^3-8)=abs(12(Deltax)+6(Deltax)^2+(Deltax)^3)#
Now chosen any #epsilon >0#, we fix #delta_epsilon# such that:
#0 < delta_epsilon < min(1, epsilon/19) #
and as for #x in (2-delta_epsilon,2+delta_epsilon)# we have #|Deltax| < delta_epsilon#
#abs (f(x) -5) = abs(12(Deltax)+6(Deltax)^2+(Deltax)^3)<= 12abs(Deltax)+6abs(Deltax)^2+abs(Deltax)^3< 12delta_epsilon+6delta_epsilon^2+delta_epsilon^3#
now, as #delta_epsilon < 1# we have #delta_epsilon^n < delta_epsilon#, so:
#abs (f(x) -5) < 12delta_epsilon+6delta_epsilon^2+delta_epsilon^3 < 12delta_epsilon+6delta_epsilon+delta_epsilon < 19delta_epsilon < epsilon#
And the limit is proved.
