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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.

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    $\begingroup$ The book is correct. The formula doesn't make sense at $x=11$. $\endgroup$ – saulspatz Jun 16 '19 at 21:46
  • $\begingroup$ The answer given by jgon is probably the best explanation for why desmos is giving you the wrong answer. $\endgroup$ – Cameron Buie Jun 16 '19 at 22:58
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    $\begingroup$ G. H. Hardy made a pretty clear statement of the principles at work here in a book review in the Mathematical Gazette, which I quoted here. $\endgroup$ – Chappers Jun 17 '19 at 12:06
  • $\begingroup$ I plotted this expression in Desmos and it clearly shows a curve approaching $(11,0)$ from below on the left and from above on the right. I zoomed in repeatedly and the only change was that the appearance of the curve became straighter. But is that really saying that Desmos evaluated the expression as zero at $x=11,$ or merely that the expression evaluates close enough to zero at points close enough to $x=11$ in order to draw all the pixels of the curve that I see? $\endgroup$ – David K Jun 17 '19 at 13:26
  • $\begingroup$ The authors of many textbooks, in order to show that there is a "hole" in a graph, will draw an empty circle around the point where the hole is. In doing this, they also erase well-defined portions of the graph near the hole. As far as I can tell, Desmos does not do any of that. It just plots all the points it can. $\endgroup$ – David K Jun 17 '19 at 13:29
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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.

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  • $\begingroup$ There are already two anonymous downvotes on this post. I don't really mind the downvotes, but if there's a way this answer could be improved, it would be helpful if you let me know. :) $\endgroup$ – jgon Jun 17 '19 at 21:41
  • $\begingroup$ OP's cited function is not a rational function, but a quotient of rational functions with natural domain $\mathbb{R}\setminus\{-2,2,11\}$. The most relevant aspects answering OP's question seems to be the reference cited in a comment by Chappers. $\endgroup$ – Markus Scheuer Jun 20 '19 at 16:09
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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 $0$).

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  • $\begingroup$ But Desmos is clearly wrong if it shows $f(11)=0$ $\endgroup$ – saulspatz Jun 16 '19 at 21:47
  • $\begingroup$ Of course it is wrong. $\endgroup$ – José Carlos Santos Jun 16 '19 at 21:49
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    $\begingroup$ I'm sorry, I didn't write the expression correctly. Please check again, thanks! $\endgroup$ – Someone7 Jun 16 '19 at 22:10
  • $\begingroup$ I've edited my answer. $\endgroup$ – José Carlos Santos Jun 16 '19 at 22:23
  • $\begingroup$ The equality holds only when the expression is defined. In other words $\frac1{\frac12}=2$ is ok, but $\frac1{\frac10}=0$ is not. $\endgroup$ – Hagen von Eitzen Jun 17 '19 at 9:54
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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.)

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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.

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

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  • $\begingroup$ Actually, $f(0)=-\frac{11}8$. $\endgroup$ – José Carlos Santos Jun 16 '19 at 21:47

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