What are the removable and non-removable discontinuities, if any, of #f(x)=(x^2 - 3x + 2)/(x^2 - 1) #?

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Answer 1 · 2015-12-10T15:13:25

Removable at #x=1# and non-removable at #x=-1#

#f(x)=(x^2 - 3x + 2)/(x^2 - 1) # is a rational function.

Therefore, #f# is continuous on its domain.

The domain of #f# is #RR - {-1,1}#.

So #f# has discontinuities at #-1# and at #1#. (It is continuous everywhere else.)

In order to determine whether a discontinuity is removable, we need to look at the limit of #f# as #x# approaches the discontinuity.

At #x=1#
#lim_(xrarr1)f(x) = lim_(xrarr1)(x^2 - 3x + 2)/(x^2 - 1)#.

This limit has indeterminate form #0/0#, so we factor and simplify.

#lim_(xrarr1)f(x) = lim_(xrarr1)((x-2)(x-1))/((x+1)(x-1)#

# = lim_(xrarr1)(x-2)/(x+1) = -1/2#

Because the limit exists, the discontinuity is removable.

At #x=-1#
#lim_(xrarr-1)f(x) = lim_(xrarr1)(x^2 - 3x + 2)/(x^2 - 1)#.

This limit has form #6/0#, so the limit does not exist and the discontinuity cannot be removed.