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 `lim_(x->a^+)=f(a)`. Function `f(x)` is continuous from the left at point `a` if `lim_(x->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`:

  1. Removable Discontinuity: if `lim_(x->a)f(x)` exists and finite, but function is either undefined at point `a` or `lim_(x->a)f(x)!=f(a)` . It is called removable because this discontinuity can be removed by redefining function as `g(x)={(f(x) if x!=a),(lim_(x->a)f(x) if x=a):}`
  2. Jump or Step Discontinuity: if one-sided limits exist and finite but not equal.
  3. 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^2+2x-3)/(x-1)` is discontinuous and classify thisremovable discontinuity and function is not defined 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, `lim_(x->1)(x^2+2x-3)/(x-1)=lim_(x->1)((x+3)(x-1))/(x-1)=`

`=lim_(x->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^2+2x-3)/(x-1) if x!=1),(2 if x=1):}` is discontinuous and classify this discontinuity.removable discontinuity and function is defined

This is actually same example as example 1, except that function is defined at `x=1`.

Again `lim_(x->1)(x^2+2x-3)/(x-1)=4`, but `f(1)=2`, so `lim_(x->1)(x^2+2x-3)/(x-1)!=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^2+2x-3)/(x-1) if x!=1),(4 if x=1):}`.

Example 3. Find where function `f(x)={(x+1 if x<=2),(4-x if x>2):}` is discontinuous and classify jump discontinuitythis discontinuity.

Clearly linear functions are continuous everywhere, but since `lim_(x->2^-)f(x)=lim_(x->2^-)(x+1)=3` and `lim_(x->2^+)f(x)=lim_(x->2^+)(4-x)=2` then `lim_(x->2^-)f(x)!=lim_(x->2^+)f(x)`.

This means that at `x=2` there is jump discontinuity.

Also note, that `lim_(x->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)=1/(x-1)` is discontinuous and classify this discontinuity.infinite 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, `lim_(x->1^+)1/(x-1)=oo`.

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: all types of discontinuities`f(x)={(1/x if x<2),(1/x if 2<x<=4),(x if 4<x<6),(4 if x=6),(x if x>6):}`

Function is discontinuous at 0 because `lim_(x->0^+)f(x)=lim_(x->0^+)1/x=oo`. So, at function `x=0` we have infinite discontinuity.

Function is dicontinuous at 2 because `f(2)` is simply not defined. Since `lim_(x->2^-)1/x=lim_(x->2^+)1/x=1/2` then `lim_(x->2)1/x=1/2` and so `x=2` is removable discontinuity.

Since `lim_(x->4^-)f(x)=lim_(x->4^-)1/x=1/4` and `lim_(x->4^+)f(x)=lim_(x->4^+)x=4` then `lim_(x->4)f(x)` doesn't exist because one-sided limits are not equal.

This means that `x=4` is jump discontinuity.

Since `lim_(x->6^-)f(x)=lim_(x->6^-)x=6` and `lim_(x->6^+)f(x)=lim_(x->6^+)x=6` then `lim_(x->6)f(x)=6` but `f(6)=4!=6` , so `lim_(x->6)f(x)!=f(6)`.

This means that `x=6` is removable discontinuity.

Example 6. Conside function `f(x)=[x]` where `[x]` is floor functioninfinite number of jump discontinuities.

This function is discontinuous at every integer point `x=n`, because `lim_(x->n^-)[x]=n-1` and `lim_(x->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=lim_(x->n^+)f(x)` that's why at every integer point `x=n` function is continuous from the right.