Two holes in the graph when dividing polynomials?

Question

Sorry if this seems like a dumb question, or if it has been answered already. I did a quick search and didn't turn up anything, so here goes.

I'm teaching a high school algebra class, and the book I'm doing says the students need to note restrictions on both the numerator and denominator when dividing two polynomials. For example:

$$\frac{ \left(\frac{x^2-121}{x^2-4} \right)}{ \left(\frac{x+2}{x-11} \right)}$$

There is obviously going to be an asymptote at $x = -2$, but the book also suggests there should be a hole in the graph at $x = 11$.

However, graphing the function on a graphing calculator (desmos, in this case), shows the function to be equal to $0$ when $x = 11$. Changing the numerator of the dividend to something like $(x² - 133)$ doesn't alter the equation equaling zero at $x = 11$ either.

Wouldn't $x = 11$ imply that $(x² - 133)/(x² - 4) ÷ (x+2)/(x-11)$ is actually $(x² - 133)/(x² - 4) ÷ 1/0$, or $(x² - 133)/(x² - 4) ÷ ∞$?

Note: the TI-83 also gives me an ERROR message for this function at $x = 11$, so perhaps it is just desmos?

Edit: Sorry! I forgot to include another division sign when discussing this problem.

Answer 1

Note that$$\frac{(x^2-121)/(x^2-4)}{(x+2)/(x-11)}=\frac{(x-11)(x+11)/(x^2-4)}{(x+2)/(x-11)}=\frac{(x+11)(x-11)^2}{(x+2)(x^2-4)}.$$Therefore,$$\lim_{x\to11}\frac{(x^2-121)/(x^2-4)}{(x+2)(x-11)}=\lim_{x\to11}\frac{(x+11)(x-11)^2}{(x+2)(x^2-4)}=0.$$In particular, this limit exists (in $\mathbb R$) and so, yes, there is a hole there ($11$ does not belong to the domain of $\frac{(x^2-121)/(x^2-4)}{(x+2)/(x-11)}$), but you can't see it if you graph the function (when $x$ is near $11$, $f(x)$ is near $0$).

Answer 2

The function is not defined at $x = 11$, so the graph does have a hole at that point.

Answer 3

I feel the need to be pedantic here, and want to make a distinction between the expression $$\frac{(x^2-121)/(x^2-4)}{(x+2)/(x-11)}$$ and the rational function it represents.

The expression is undefined at $x=11$, since we cannot evaluate $(x+2)/(x-11)$ when $x=11$.

However the rational function it represents can also be written as $$\frac{(x^2-121)(x-11)}{(x+2)(x^2-4)},$$ so the rational function is defined at $x=11$, because it can be written as a fraction of polynomials for which $11$ is not a root of the polynomial in the denominator.

There are several good reasons to make this distinction, though I'm not sure it's usually made at a high school level, but I'll cite Wikipedia as a source to show that this distinction is in fact made.

An aside on why this distinction is made

The short version is that algebraically, we can think of rational functions as fractions of polynomials. Thus if two fractions are equal, then the functions they define should also be equal.

I.e., because these fractions of polynomials are equal, $$\frac{1}{x-2} = \frac{x+2}{x^2-4}$$ they should represent the same function.

This is related to the idea that polynomials aren't functions. See this question for more on that topic.

Answer 4

There are three divisions in the recipe you have written, so there are three opportunities for division by zero. If $x = \pm 2$, the fraction in the numerator is a division by zero, so is undefined. If $x = 11$, the fraction in the denominator is division by zero, so is undefined. If $x = -2$, the denominator is zero, so the "big" fraction is undefined. This puts "holes" in the graph at $-2$, $2$, and $11$.

Desmos is applying the identity $$ \frac{\, \frac{a}{b}\, }{\frac{c}{d}} = \frac{a}{b} \cdot \frac{d}{c} \text{,} $$ so "forgets" the division by zero when $x = 11$ because it has forgotten that $d$ was a denominator. This is incorrect behaviour by Desmos. An identity is only required to be an equality of values when both expressions (one on each side of the equality) are defined. As already explained, the recipe you wrote is not defined at $x = 11$, so the two sides of the identity need not agree at $x = 11$. (And they do not : equality is a relation between values and "undefined" is not a value. The easiest example of this is "$1 = \dfrac{x}{x}$", which is an equality of values for $x \neq 0$ and, since at least one side is undefined at $x = 0$, need not be an equality at $x = 0$ for the identity to be valid.)

Answer 5

The hint of @Chappers is valuable and deserves an answer by its own.

OP's stated function \begin{align*} \frac{ \left(\frac{x^2-121}{x^2-4} \right)}{ \left(\frac{x+2}{x-11} \right)} \end{align*} is a quotient of rational functions with natural domain $\mathbb{R}\setminus\{-2,2,11\}$ which differs from the rational function \begin{align*} \frac{(x^2-121)(x-11)}{(x+2)(x^2-4)} \end{align*} with natural domain $\mathbb{R}\setminus\{-2,2\}$.

The following examples given by G.H. Hardy in Mathematical Gazette 1907, 4 pp. 13–14 might help to clarify the situation.

• The function $\frac{x^2-1}{x-1}$ has no value for $x=1$; for $x=1$, $\frac{x^2-1}{x-1}$ is strictly and absolutely meaningless. The fact that its limit for $x=1$ is $2$ is entirely irrelevant. The functions $\frac{x^2-1}{x-1}$ and $x+1$ are different functions. They are equal when $x$ is not equal to $1$.

• Similarly the function $y=\frac{x}{x}$ is $=1$ when $x \neq 0$ and undefined for $x=0$. To calculate $f(x)$ for $x=0$ we must put $x=0$ in the expression of $f(x)$ and perform the arithmetical operations which the form of the function prescribes, and this we cannot do in this case.

Answer 6

I see three denominators in the expression $\frac{ \left(\frac{x^2-121}{x^2-4} \right)}{ \left(\frac{x+2}{x-11} \right)}$

• $x^2-4$
• $x-11$
• $\frac{x+2}{x-11}$

None of them can be allowed to be equal to $0$.

Sometimes, simplifying can make a discontinuity go away. It doesn't matter. You have to deal with the original expression.

That means that the expression is undefined when $x \in \{2, -2, 11\}$