Let #f(x)=(x^3 - 4x)/(x^3 +x^2-6x)#, how do you find all points of discontinuity of f(x)?

#f(x) = (x^3-4x)/(x^3+x^2-6x) = (x(x-2)(x+2))/(x(x-2)(x+3)) = (x+2)/(x+3)#

with exclusions #x != 0# and #x != 2#.

There are removable discontinuities at #x=0# and #x=2# where both the numerator and the denominator are zero, so #f(x)# is undefined at those points, but the left and right limits agree.

There is a simple pole at #x=-3#, where the denominator is zero and the numerator is non-zero.

#lim_(x->-3+) f(x) = -oo#
#lim_(x->-3-) f(x) = +oo#

Apart from these discontinuities, #f(x)# is well defined and continuous.

graph{(x+2)/(x+3) [-10, 10, -5, 5]}

A rational function is continuous on its domain.

So the points of discontinuity for a rational function are the point outside the domain.

#x^3+x^2-6x = 0#

#x(x^2+x-6) = x(x-2)(x+3) = 0#

The points outside the domain are: #0, " "2, " and "-3#

Note

Because this was posted in the topic "Classifying Discontinuities, I should probably add that

#f(x) = (x(x+2)(x-2))/(x(x-2)(x+3)) = (x+2)/(x+3)# for #x != 0, 2#

So the only infinite limit occurs at #x=-3#. There is a non-removable (infinite) discontinuity at #-3#

The discontinuities at #0# and #2# are removable.