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

2 Answers
Nov 1, 2015

Removable discontinuity at x=3.
No non-removable discontinuity.

Explanation:

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

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

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

We have the only discontinuity at x=3 and it is removable.

Explanation:

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.