Function #f# is continuous at #a# if and only if
#lim_(xrarra)f(x) = f(a)#.
There are three conditions in this equality:
#C1:# #f(a)# exists
#C2:# #lim_(xrarra)f(x)# exists
#C3:# The numbers in #C1# and #C2# are the same.
#f(x) = (x-1)/(x^2+x-2)# is not defined for solutions to #x^2+x-2=0#.
Therefore, #f# is not continuous at #x=1# and at #x=-2#
A discontinuity at #a# is removable if the function is discontinuous at #a#, but the limit as #x# approaches #a# exists.
Checking the two discontinuities we found, we see that
#lim_(xrarr1)f(x) = lim_(xrarr1)(x-1)/((x-1)(x+2)) = lim_(xrarr1)1/(x+2) = 1/3#
Because this limit exists, the discontinuity at #1# is removable.
#lim_(xrarr-2)f(x)# does not exist. SO the discontinuity at #-2# is not removable. (In fact, it is an infinite discontinuity)
To remove the discontinuity at #1#, define a new function, #g# by
#g(x) = {(f(x)," if ",x != 1" and " x!= -2),(1/3," if ",x=1):}#.
#g(x) = f(x)# for all #x# except #1# (and #-2#) and #g# is continuous at #x=1#.
