Discontinuities

We take a look at the various types of discontinuity and how they occur.

Removable Discontinuity

A removable discontinuity is a point $\ x=c$ where the function has a discontinuity, but may be redefined at that point to make it continuous.

\begin{displaymath}
f(x)=\frac{x^{2}-1}{x-1}
\end{displaymath}


You can see that this function has a discontinuity at $x=1$ (in fact this point is not in the domain of $f$ ) however if we define $f(1)=2$ , then this becomes a continuous function.

Jump Discontinuity

This is where the function has a jump either side of the point $\ x=c$ .


The function $f(x)$ has the given graph. It has a jump discontinuity at $x=0$. Notice that there is no way to redefine $f$ at $x=0$ to make it continuous as in the previous example.


Infinite Discontinuity

A function has an infinite discontinuity if the limit at $\ x=c$ is plus or minus infinity. $f(x)=1/x^{2}$


$f(x)=1/x$


Oscillatiing Discontinuity

An oscillating discontinuity occurs when the value of the function is changing so rapidly that a limit is not possible. The classic example is

\begin{displaymath}
f(x)=\sin (1/x)
\end{displaymath}



Exercises

On what intervals are the following functions continuous? Beware of removable discontinuities which this program will ignore!

  1. \begin{displaymath}
f(x)=\frac{x+1}{x^{2}-4x+3}
\end{displaymath}


  2. \begin{displaymath}
f(x)=\frac{1}{\vert x\vert+1}-\frac{x^{2}}{2}
\end{displaymath}


  3. \begin{displaymath}
f(x)=\frac{\cos x}{x}
\end{displaymath}

    Are the following functions continuous at the given point?
  4. $\sin (x=\sin x)$ at $x=\pi $
  5. $\sin (\frac{\pi }{2}\cos (\tan x))$ at $x=0$
  6. $\sec (x\sec ^{2}x-\tan ^{2}x-1)$ at $x=1$




Robert Judd
2001-11-24