What are the removable and non-removable discontinuities, if any, of #f(x)=(3x^2-7x-6 )/ (x-3 ) #?

#f(x)=frac{3x^2-7x-6}{x-3}#

#=(3x+2)frac{x-3}{x-3}#

#lim_{x->3}f(x)=8#

When dealing with quotients of continuous functions, discontinuities arise at points that nullify the denominator.
In our specific case #f(x)=(3x^2-7x-6)/(x-3)#, so the denominator is null if #x=3#. Then the domain of this function is the union of two open intervals #D=]-infty,3[ cup ]3,+infty[# and at #x=3# we have a discontinuity.

Intuitively, a discontinuity is removable if we can extend the function to a continuous one by assigning a value to the point where the discontinuity occurs. More formally, if #x_0# is a discontinuity of #f:]a,x_0[ cup ]x_0,b[ to RR#, the discontinuity is said to be removable if exists finite #lim_{x to x_ 0}f(x)#.

In our specific case we want to check if #lim_{x to 3}(3x^2-7x-6)/(x-3)# exists and if it's finite. Since #3# is a root of the numerator, we conclude that #3x^2-7x-6# can be factored out by #x-3#. We have that #3x^2-7x-6=(3x+2)(x-3)#, so
#lim_{x to 3}(3x^2-7x-6)/(x-3)=lim_{x to 3}((3x+2)(x-3))/(x-3)=lim_{x to 3}(3x+2)=11#

Since the limit exists and is finite, #x=3# is a removable discontinuity.