One-Sided Continuity. Classification of Discontinuities
Similarly to the one-sided limits, we can define one-sided continuity.
Definition. Function f(x) is continuous from the right at point a if limx→a+=f(a). Function f(x) is continuous from the left at point a if limx→a−f(x)=f(a).
Clearly, if function is continuous from the left and from the right at point a, then it is continuous at point a.
Definition. Function f is discontinuous at a if it is not continuous.
There are three kinds of discontinuity at a:
Removable Discontinuity: if limx→af(x) exists and finite, but function is either undefined at point a or limx→af(x)=f(a). It is called removable because this discontinuity can be removed by redefining function as g(x)={f(x)ifx=alimx→af(x)ifx=a
Jump or Step Discontinuity: if one-sided limits exist and finite but not equal.
Infinite or Essential Discontinuity: if one or both of the one-sided limits don't exist or are infinite.
Let's go through a couple of examples of these discontinuities.
Example 1. Find where function f(x)=x−1x2+2x−3 is discontinuous and classify this discontinuity.
This function is rational, so it is continuous everywhere, except where denominator equals 0, i.e. where x−1=0. So, functions is not defined (and not continuous) when x=1.
Now, limx→1x−1x2+2x−3=limx→1x−1(x+3)(x−1)=
=limx→1(x+3)=4.
Thus, limit exists and finite, but f(1) is not defined, so x=1 is removable discontinuity.
Example 2. Find where function f(x)={x−1x2+2x−3ifx=12ifx=1 is discontinuous and classify this discontinuity.
This is actually same example as example 1, except that function is defined at x=1.
Again limx→1x−1x2+2x−3=4, but f(1)=2, so limx→1x−1x2+2x−3=f(1).
Thus, x=1 is point of removable discontinuity.
As in example 1 we can make function continuous by redefining function as f(x)={x−1x2+2x−3ifx=14ifx=1.
Example 3. Find where function f(x)={x+1ifx≤24−xifx>2 is discontinuous and classify this discontinuity.
Clearly linear functions are continuous everywhere, but since limx→2−f(x)=limx→2−(x+1)=3 and limx→2+f(x)=limx→2+(4−x)=2 then limx→2−f(x)=limx→2+f(x).
This means that at x=2 there is jump discontinuity.
Also note, that limx→2−f(x)=3=f(1), this means that function is continuous from the left at x=2.
Example 4. Find where function f(x)=x−11 is discontinuous and classify this discontinuity.
This function is rational, so it is continuous everywhere, except where denominator equals 0, i.e. where x−1=0. So, functions is not continuous when x=1.
Now, limx→1+x−11=∞.
This means that at x=1 there is infinite discontinuity.
Actually infinite discontinuity occurs when we have vertical asymptote.
Example 5. Find points where function f is discontinuous and classify these points: f(x)=⎩⎨⎧x1ifx<2x1if2<x≤4xif4<x<64ifx=6xifx>6
Function is discontinuous at 0 because limx→0+f(x)=limx→0+x1=∞. So, at function x=0 we have infinite discontinuity.
Function is dicontinuous at 2 because f(2) is simply not defined. Since limx→2−x1=limx→2+x1=21 then limx→2x1=21 and so x=2 is removable discontinuity.
Since limx→4−f(x)=limx→4−x1=41 and limx→4+f(x)=limx→4+x=4 then limx→4f(x) doesn't exist because one-sided limits are not equal.
This means that x=4 is jump discontinuity.
Since limx→6−f(x)=limx→6−x=6 and limx→6+f(x)=limx→6+x=6 then limx→6f(x)=6 but f(6)=4=6, so limx→6f(x)=f(6).
This means that x=6 is removable discontinuity.
Example 6. Conside function f(x)=[x] where [x] is a floor function.
This function is discontinuous at every integer point x=n, because limx→n−[x]=n−1 and limx→n+[x]=n.
So one-sided limits are finite but not equal. This means that at every integer point x=n there is jump discontinuity.
Also, note that f(n)=n=limx→n+f(x) that's why at every integer point x=n function is continuous from the right.