Question

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

Answer 1

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`

Answer 2

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.