Removable Discontinuity: A removable discontinuity is a point on the graph that is undefined or does not fit the rest of the graph. There is a gap at that location when you are looking at the graph. Rational functions are discontinuous when the denominator equals zero. To find the removable discontinuity:

How to Find:
1. Factor the numerator and the denominator
2. Identify factors that occur in both the numerator and the denominator
3. Set the common factors equal to zero
4. Solve for x
5. Write your answers in the form x=
6. To find t=y plug the x values in the remaining parts of the equation


\begin{align} f(x)=\frac{x}{(x+2)(x-3)}\ \end{align}

(has discontinuities at x=-2 and at x=3 so the denominator would be equal to zero.)


\begin{align} f(x)=\frac{8x^2+26x+15}{2x^2-x-15}\ \Rightarrow\frac{(4x+3)(2x+5)}{(2x+5)(x-3)}\ \end{align}

Solve for x:

\begin{align} 2x+5 \Rightarrow x=\frac{-5}{2}\ \end{align}

Solve for y:

\begin{align} \frac{4(\frac{-5}{2})+3}{\frac{-5}{2}-3} \Rightarrow \frac{-10+3}{\frac{-11}{2}}\ \Rightarrow \frac{-7}{\frac{-11}{2}} \Rightarrow y=\frac{14}{11}\ \end{align}

Removable Discontinuity at

\begin{align} (\frac{-5}{2}\ \textrm{,} \frac{14}{11})\ \end{align}

Jump Discontinuity: If the left and right hand limits at x=a exist but disagree, then the graph jumps at x=a.

The point x=2 is a Jump Discontinuity.
Infinite Discontinuity: Exists when one of the one-sided limits of the function is infinite.Example: The graph below shows a function that is discontinuous at x=a
Since the function never reaches a final value, the limit does not exist.