We have seen that for some fractional functions, there are values for which the numerator and the denominator both equal zero. In some cases, such values are perforations of the function. Calculating them gives us an insight into what the function looks like without having to draw the complete graph of the function. We first give a general definition of a perforation.
Function #f(x)# has a perforation at #x=\orange{a}# if #\phantom{\frac{g(x)}{h(x)}}#\[\lim_{x\downarrow \orange{a}} f(x) = \lim_{x\uparrow \orange{a}} f(x) \] both limits are finite and #f(\orange{a})# is not defined.
Example
Let #f(x)=\frac{x^2+x-2}{x+2}#. We have \[\lim_{x\downarrow \orange{-2}} f(x) = -3 \text{ and } \lim_{x\uparrow \orange{-2}} f(x) = -3\] and the denominator of #f(x)# is zero at #x=\orange{-2}#, so #f(\orange{a})# is not defined. Thus, #f(x)# has a perforation at #x=\orange{a}#.
In different literature, a perforation is sometimes referred to as removable discontinuity, even though the definition of removable discontinuity is an extension of that of a perforation. Further theory on (dis)continuity is outside the scope of this course.
If #\lim_{x\downarrow \orange{a}} f(x) = \lim_{x\uparrow \orange{a}} f(x)# and #f(\orange{a})#
is defined, then there are two possibilities. Either \[\lim_{x\downarrow \orange{a}} f(x) = \lim_{x\uparrow \orange{a}} f(x)= \lim_{x\to \orange{a}} f(x)=f(\orange{a}),\] like the example on the left, or \[\lim_{x\downarrow \orange{a}} f(x) = \lim_{x\uparrow \orange{a}} f(x)\neq \lim_{x\to \orange{a}} f(x),\] like the example on the right.
Example
#f(x)=x# at #x=\orange{1}#. #\phantom{\begin{cases} i\\ i \end{cases}}#
Example
#f(x)=\begin{cases} x & x\neq 1\\ 0 & x=1 \end{cases}# at #x=\orange{1}#.
In neither of these cases we are dealing with a perforation. However, we are dealing with a removable discontinuity in the right case.
We need the condition that #\lim_{x\downarrow \orange{a}} f(x) = \lim_{x\uparrow \orange{a}} f(x)# are finite. If they are infinite instead, then we are dealing with a
vertical asymptote at #x=\orange{a}# instead of a perforation.
For fractional functions, perforations are quite easy to find.
Let #f(x)=\frac{\blue{g(x)}}{\green{h(x)}}#. If there is a value #\orange{a}# such that \[\blue{g(}\orange{a}\blue{)}=\green{h(}\orange{a}\green{)}=0\] then #f(x)# has a perforation at #x=\orange{a}#.
Example
Let again #f(x)=\frac{\blue{g(x)}}{\green{h(x)}}=\frac{\blue{x^2+x-2}}{\green{x+2}}#. Since #x^2+x-2=(x+2)(x-1)#, we have that \[\blue{g(}\orange{-2}\blue{)}=\green{h(}\orange{-2}\green{)}=0,\] so #f(x)# has a perforation at #x=\orange{-2}#.
Note that if #\orange{a}# is a perforation of #f(x)#, then the denominator of #f(x)# equals zero at #x=\orange{a}#, and thus #f(x)# is indeed not defined at #x=\orange{a}#.
The graph of the function #f(x)=\frac{\blue{g(x)}}{\green{h(x)}}=\frac{\blue{x^2+x-2}}{\green{x+2}}# is displayed in the following figure. We see that indeed a perforation occurs in #x=\orange{-2}#.
Because of the requirement that #\lim_{x\downarrow \orange{a}} f(x) = \lim_{x\uparrow \orange{a}} f(x)# need to be finite, there are some cases that have #\blue{g(}\orange{a}\blue{)}=\green{h(}\orange{a}\green{)}=0#, but where a vertical asymptote occurs at #x=\orange{a}# instead of a perforation. These kinds of functions are outside the scope of this course.
When the left and right limits at a certain point are not equal, we are dealing with a different kind of discontinuity.
Function #f(x)# has a jump discontinuity at #x=\orange{a}# if \[\lim_{x\downarrow \orange{a}} f(x) \neq \lim_{x\uparrow \orange{a}} f(x) \] and both #\lim_{x\downarrow \orange{a}} f(x)# and #\lim_{x\uparrow \orange{a}} f(x)# are finite.
A jump discontinuity can also be called a step discontinuity or a discontinuity of the first kind.
There are three types of discontinuities. We have seen two of them already, namely perforations (also known as removable discontinuities) and jump discontinuities. The third type of discontinuities is called essential discontinuities.
A function #f(x)# has an essential discontinuity in #x=\orange{a}# if at least one of #\lim_{x\downarrow \orange{a}} f(x)# or #\lim_{x\uparrow \orange{a}} f(x)# is infinite.
An essential discontinuity can also be called an
infinite discontinuity or a
discontinuity of the second kind.
Consider the function \[f(x)={{x^3+8 x ^2-36 x -288}\over{x^3+23 x ^2+174 x +432}}\] Find all perforations of this function, if there are any.
The function #f(x)# has perforation in #x=-6# and #x=-8#.
The answer can be found as follows.
\[\begin{array}{rcl}
\displaystyle f(x) &=& \displaystyle {{x^3+8 x ^2-36 x -288}\over{x^3+23 x ^2+174 x +432}} \\
&& \qquad \blue{\text{function rule for } f(x) \text{ used}}\\
&=& \displaystyle \frac{(x+6)\cdot(x+8)\cdot (x-6)}{(x+6)\cdot(x+8)\cdot(x+9)}\\
&& \qquad\blue{\text{numerator and denominator factored}}
\end{array}\]
We see that the numerator equals zero at #x=-6#, #x=-8# and #x=6#. The denominator equals zero at #x=-6#, #x=-8# and #x=-9#. Thus, both numerator and denominator equal zero at #x=-6# and #x=-8#. This means #f(x)# has two perforations, namely at #x=-6# and #x=-8#.
Perforations and other discontinuities
