Question

How do you find the x values at which `f(x)=abs(x-3)/(x-3)` is not continuous, which of the discontinuities are removable?

Answer 1

Refer to the Discussion given in the Explanation Section below.

Explanation:

Observe that `f` is not defined at `x=3`, and, hence is not

continuous at that point.

For `AA x in (3,oo) ={ x in RR : x>3};` by the defn. of Absolute Value Function,

`|x-3|=(x-3) rArr f(x)=|x-3|/(x-3)=(x-3)/(x-3)=1, x >3.`

`"Similarly, "AA x in (-oo,3), f(x)=(-(x-3))/(x-3)=-1, x<3.`

This means that `lim_(x to 3+) f(x)=1 != -1 = lim_(x to 3-) f(x).`

In other words, `lim_(x to 3) f(x)" does not exist."`

We conclude that `f" has non-removable discontinuity at "x=3.`

Answer 2

`f(x)` is continuous in `RR -{3}`, and the dicontinuity is not removable.

Explanation:

The function is discontinuous for `x=3` since the denominator is null.

Then we analyze separately:

1. For `x in (3,+oo)`, we have:
   `(x-3) > 0 => abs(x-3)= (x-3) => f(x) = abs(x-3)/(x-3)= (x-3)/(x-3) = 1`

2. For `x in (-oo,3)`, we have:
   `(x-3) < 0 => abs(x-3)= -(x-3) => f(x) = abs(x-3)/(x-3)= -(x-3)/(x-3) = -1`

Thus, `f(x)` is continuous in `RR -{3}`

The discontinuity cannot be removed, as clearly:

`lim_(x->3^-) f(x) = -1`

`lim_(x->3^+) f(x) = 1`

As the limit from the right and from the left are different, then the function does not have a limit for `x->3`

This is the graph of `y=f(x)` which shows the discontinuity:
graph{(|x-3|)/(x-3) [-2, 5, -3, 3]}