One-Sided Continuity. Classification of Discontinuities

Similarly to the one-sided limits, we can define one-sided continuity.

Definition. Function f(x){f{{\left({x}\right)}}} is continuous from the right at point a{a} if limxa+=f(a)\lim_{{{x}\to{{a}}^{+}}}={f{{\left({a}\right)}}}. Function f(x){f{{\left({x}\right)}}} is continuous from the left at point a{a} if limxaf(x)=f(a)\lim_{{{x}\to{{a}}^{{-}}}}{f{{\left({x}\right)}}}={f{{\left({a}\right)}}}.

Clearly, if function is continuous from the left and from the right at point a{a}, then it is continuous at point a{a}.

Definition. Function f{f{}} is discontinuous at a{a} if it is not continuous.

There are three kinds of discontinuity at a{a}:

  1. Removable Discontinuity: if limxaf(x)\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}} exists and finite, but function is either undefined at point a{a} or limxaf(x)f(a)\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}\ne{f{{\left({a}\right)}}}. It is called removable because this discontinuity can be removed by redefining function as g(x)={f(x)ifxalimxaf(x)ifx=a{g{{\left({x}\right)}}}={\left\{\begin{array}{c}{f{{\left({x}\right)}}}{\quad\text{if}\quad}{x}\ne{a}\\\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}{\quad\text{if}\quad}{x}={a}\\ \end{array}\right.}
  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)=x2+2x3x1{f{{\left({x}\right)}}}=\frac{{{{x}}^{{2}}+{2}{x}-{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 x1=0{x}-{1}={0}. So, functions is not defined (and not continuous) when x=1{x}={1}.

Now, limx1x2+2x3x1=limx1(x+3)(x1)x1=\lim_{{{x}\to{1}}}\frac{{{{x}}^{{2}}+{2}{x}-{3}}}{{{x}-{1}}}=\lim_{{{x}\to{1}}}\frac{{{\left({x}+{3}\right)}{\left({x}-{1}\right)}}}{{{x}-{1}}}=

=limx1(x+3)=4=\lim_{{{x}\to{1}}}{\left({x}+{3}\right)}={4}.

Thus, limit exists and finite, but f(1){f{{\left({1}\right)}}} is not defined, so x=1{x}={1} is removable discontinuity.

Example 2. Find where function f(x)={x2+2x3x1ifx12ifx=1{f{{\left({x}\right)}}}={\left\{\begin{array}{c}\frac{{{{x}}^{{2}}+{2}{x}-{3}}}{{{x}-{1}}}{\quad\text{if}\quad}{x}\ne{1}\\{2}{\quad\text{if}\quad}{x}={1}\\ \end{array}\right.} 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{x}={1}.

Again limx1x2+2x3x1=4\lim_{{{x}\to{1}}}\frac{{{{x}}^{{2}}+{2}{x}-{3}}}{{{x}-{1}}}={4}, but f(1)=2{f{{\left({1}\right)}}}={2}, so limx1x2+2x3x1f(1)\lim_{{{x}\to{1}}}\frac{{{{x}}^{{2}}+{2}{x}-{3}}}{{{x}-{1}}}\ne{f{{\left({1}\right)}}}.

Thus, x=1{x}={1} is point of removable discontinuity.

As in example 1 we can make function continuous by redefining function as f(x)={x2+2x3x1ifx14ifx=1{f{{\left({x}\right)}}}={\left\{\begin{array}{c}\frac{{{{x}}^{{2}}+{2}{x}-{3}}}{{{x}-{1}}}{\quad\text{if}\quad}{x}\ne{1}\\{4}{\quad\text{if}\quad}{x}={1}\\ \end{array}\right.}.

Example 3. Find where function f(x)={x+1ifx24xifx>2{f{{\left({x}\right)}}}={\left\{\begin{array}{c}{x}+{1}{\quad\text{if}\quad}{x}\le{2}\\{4}-{x}{\quad\text{if}\quad}{x}>{2}\\ \end{array}\right.} is discontinuous and classify jump discontinuitythis discontinuity.

Clearly linear functions are continuous everywhere, but since limx2f(x)=limx2(x+1)=3\lim_{{{x}\to{{2}}^{{-}}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{2}}^{{-}}}}{\left({x}+{1}\right)}={3} and limx2+f(x)=limx2+(4x)=2\lim_{{{x}\to{{2}}^{+}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{2}}^{+}}}{\left({4}-{x}\right)}={2} then limx2f(x)limx2+f(x)\lim_{{{x}\to{{2}}^{{-}}}}{f{{\left({x}\right)}}}\ne\lim_{{{x}\to{{2}}^{+}}}{f{{\left({x}\right)}}}.

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

Also note, that limx2f(x)=3=f(1)\lim_{{{x}\to{{2}}^{{-}}}}{f{{\left({x}\right)}}}={3}={f{{\left({1}\right)}}}, this means that function is continuous from the left at x=2{x}={2}.

Example 4. Find where function f(x)=1x1{f{{\left({x}\right)}}}=\frac{{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 x1=0.{x}-{1}={0}. So, functions is not continuous when x=1{x}={1}.

Now, limx1+1x1=\lim_{{{x}\to{{1}}^{+}}}\frac{{1}}{{{x}-{1}}}=\infty.

This means that at x=1{x}={1} there is infinite discontinuity.

Actually infinite discontinuity occurs when we have vertical asymptote.

Example 5. Find points where function f{f{}} is discontinuous and classify these points: all types of discontinuitiesf(x)={1xifx<21xif2<x4xif4<x<64ifx=6xifx>6{f{{\left({x}\right)}}}={\left\{\begin{array}{c}\frac{{1}}{{x}}{\quad\text{if}\quad}{x}<{2}\\\frac{{1}}{{x}}{\quad\text{if}\quad}{2}<{x}\le{4}\\{x}{\quad\text{if}\quad}{4}<{x}<{6}\\{4}{\quad\text{if}\quad}{x}={6}\\{x}{\quad\text{if}\quad}{x}>{6}\\ \end{array}\right.}

Function is discontinuous at 0 because limx0+f(x)=limx0+1x=\lim_{{{x}\to{{0}}^{+}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{1}}{{x}}=\infty. So, at function x=0{x}={0} we have infinite discontinuity.

Function is dicontinuous at 2 because f(2){f{{\left({2}\right)}}} is simply not defined. Since limx21x=limx2+1x=12\lim_{{{x}\to{{2}}^{{-}}}}\frac{{1}}{{x}}=\lim_{{{x}\to{{2}}^{+}}}\frac{{1}}{{x}}=\frac{{1}}{{2}} then limx21x=12\lim_{{{x}\to{2}}}\frac{{1}}{{x}}=\frac{{1}}{{2}} and so x=2{x}={2} is removable discontinuity.

Since limx4f(x)=limx41x=14\lim_{{{x}\to{{4}}^{{-}}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{4}}^{{-}}}}\frac{{1}}{{x}}=\frac{{1}}{{4}} and limx4+f(x)=limx4+x=4\lim_{{{x}\to{{4}}^{+}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{4}}^{+}}}{x}={4} then limx4f(x)\lim_{{{x}\to{4}}}{f{{\left({x}\right)}}} doesn't exist because one-sided limits are not equal.

This means that x=4{x}={4} is jump discontinuity.

Since limx6f(x)=limx6x=6\lim_{{{x}\to{{6}}^{{-}}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{6}}^{{-}}}}{x}={6} and limx6+f(x)=limx6+x=6\lim_{{{x}\to{{6}}^{+}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{6}}^{+}}}{x}={6} then limx6f(x)=6\lim_{{{x}\to{6}}}{f{{\left({x}\right)}}}={6} but f(6)=46{f{{\left({6}\right)}}}={4}\ne{6}, so limx6f(x)f(6)\lim_{{{x}\to{6}}}{f{{\left({x}\right)}}}\ne{f{{\left({6}\right)}}}.

This means that x=6{x}={6} is removable discontinuity.

Example 6. Conside function f(x)=[x]{f{{\left({x}\right)}}}={\left[{x}\right]} where [x]{\left[{x}\right]} is a floor functioninfinite number of jump discontinuities.

This function is discontinuous at every integer point x=n{x}={n}, because limxn[x]=n1\lim_{{{x}\to{{n}}^{{-}}}}{\left[{x}\right]}={n}-{1} and limxn+[x]=n\lim_{{{x}\to{{n}}^{+}}}{\left[{x}\right]}={n}.

So one-sided limits are finite but not equal. This means that at every integer point x=n{x}={n} there is jump discontinuity.

Also, note that f(n)=n=limxn+f(x){f{{\left({n}\right)}}}={n}=\lim_{{{x}\to{{n}}^{+}}}{f{{\left({x}\right)}}} that's why at every integer point x=n{x}={n} function is continuous from the right.