Holes of Rational Functions (College Board AP® Precalculus): Revision Note
Holes of rational functions
What is a hole of a rational function?
A hole is sometimes represented by an open circle on a graph
For a rational function, it corresponds to an input value for which the function is not defined
Unlike at a vertical asymptote, however, the function does not display unbounded behavior on either side of a hole

Examiner Tips and Tricks
Be careful when using a graphing calculator, as holes in a rational function will probably not be displayed or be otherwise visually obvious. This is different from a vertical asymptote, where the unbounded behavior of the function on either side of the asymptote is obvious.
Where do holes occur in rational functions?
Let be a rational function
i.e. where and are both polynomial functions
If is a real zero of both and , i.e. if and
and if the multiplicity of as a real zero of is greater than or equal to its multiplicity as a real zero of
then the graph of has a hole at
E.g. and both have holes at
because in both cases the numerator and denominator become zero when
and the multiplicity of in the numerator (4 or 7) is greater than or equal to its multiplicity in the denominator (4)
Note that the expression can be simplified to give
and the expression can be simplified to give
If the multiplicity of as a real zero of is less than its multiplicity as a real zero of , then there is a vertical asymptote at
Examiner Tips and Tricks
The expression for a rational function with a hole at will always be able to be simplified to an expression where the denominator is not equal to zero when . This is a simple way to distinguish a hole from a vertical asymptote.
How do I express the behavior of rational functions near holes using limit notation?
If a rational function has a hole at
then for input values near the hole the output values of the function will converge towards a particular value
The value they converge to determines the -coordinate of the hole
E.g. consider the behavior of close to
The output value gets closer and closer to 1.75 as gets closer and closer to 3
2.9 | 1.76923076923 |
2.99 | 1.75187969925 |
2.999 | 1.75018754689 |
2.9999 | 1.75001875047 |
3 | undefined |
3.0001 | 1.74998125047 |
3.001 | 1.74981254686 |
3.01 | 1.74812967581 |
3.1 | 1.73170731707 |
If output values of become arbitrarily close to some number as input values become sufficiently close to
then the hole occurs at the point
and the corresponding limit notation is
can be read as "the limit as approaches "
Examiner Tips and Tricks
Note that in the case of a hole at
I.e. the limits 'from the left' and 'from the right' are both equal to the same number
How can I determine the limit of a rational function at a hole?
One way to determine the limit at a point where a rational function has a hole
is to calculate output values of the function nearer and nearer to the hole
and see what value they seem to converge towards
For example, see the table of values for above
However a simpler method can be to algebraically simplify the expression for the function
and then find the value of the simplified expression at the input value where the hole occurs
E.g. you can see from the table above that
Instead, start by simplifying the expression
Substitute into the simplified expression
That can be a much quicker way to find the limit (especially on a non-calculator section of the exam)
unless a question specifies that you must use another method
Examiner Tips and Tricks
Be careful here. The expression can be simplified algebraically to . However
and
are not 'the same function'!
Their behaviors differ at the point where
has a well-defined value at , i.e.
is undefined at , and has a hole at
Worked Example
In the -plane, the graph of a rational function has a hole at . Which of the following could be an expression for ?
(A)
(B)
(C)
(D)
Answer:
A hole can occur where an input value makes both the numerator and denominator of a rational function zero
When ,
So options (B) and (D) will both have numerators and denominators equal to zero when
But it is only a hole if simplifying the rational expression can make the zero in the denominator go away
In option (D),
That simplified form still has a denominator equal to zero when
but the numerator is no longer equal to zero
so is a vertical asymptote, not a hole
In option (B), the rational expression can be simplified to
The denominator of the simplified form is not equal to zero when
So is a hole of
(B)
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