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I'm reading Larson's AP Calculus textbook and in the section on limits (1.3) it suggests finding functions that "agree at all but one point" in order to evaluate limits analytically. For example, given $$ f(x) = \frac{x^3-1}{x-1} $$ we can factor and reduce to get $$ \frac{x^3-1}{x-1} = \frac{(x-1)(x^2+x+1)}{x-1} = \frac{\require{cancel} \bcancel{(x-1)}(x^2+x+1)}{\require{cancel} \bcancel{(x-1)}}=x^2+x+1 $$ But it then refers to this reduced expression as $$ g(x) = x^2+x+1 $$ suggesting that $f(x) \neq g(x)$ (also implied by the figure below).

enter image description here

Is it true that $f(x) \neq g(x)$? I realize that $f(x)$ as originally expressed is indeterminate at $x=1$, but doesn't the cancellation of $(x-1)$ allow us to calculate the "true" value of $f(1)$? If so, I believe the graph of $f(x)$ in the figure is misleading at best by drawing the plot as being undefined at $x=1$.

Carser
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  • If $\lim_{x\to a}f(x)$ exists, and $f(x)=g(x)$ for all $x\ne a$, then by the definition of a limit, it shouldn't matter whether or not $f(x)$ truly equals $g(x)$ or not, as they are equal on enough of a domain to take the limit, which is the ultimate goal here. Of course, we must also have $g(x)$ be continuous around $x=a$ and be defined for $x=a$ so that we may easily use it for the limit. – Simply Beautiful Art Jul 27 '16 at 13:04
  • Take not that one of the first topics on limits was $\lim_{x\to a}f(x)\ne f(a)$ always. (Painful piece-wise functions) So I don't think you could call it the "true" value of $f(1)$ – Simply Beautiful Art Jul 27 '16 at 13:07
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    There's no such thing as an expression being "indeterminate". Limits can be indeterminate (which doesn't mean that the limit doesn't exist, only that a particular method for finding it won't work), but your function is not defined using limits at all. However, the expression $\frac{x^3-1}{x-1}$ has no value when $x=1$ (this is different from being "indeterminate") and if you want to define a function such that $f(1)=3$, then defining it as $\frac{x^3-1}{x-1}$ will not suffice. You need either an explicit special case, or an explicit convention of ignoring removable singularities. – hmakholm left over Monica Jul 27 '16 at 13:08
  • @HenningMakholm that makes so much more sense! :) – Carser Jul 27 '16 at 13:49

2 Answers2

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The first function is not defined when $x = 1$, whereas the second is. The cancelling is only possible assuming that $x \neq 1$; when $x = 1$ the denominator is zero and hence the expression is undefined.

Since the domains are different, the functions are different. However, the functions are obviously equal when restricted to the domain that excludes $x = 1$.

Eff
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To reiterate what Eff said, a function is more than a formula; it requires a domain. The domain of $f$ and the domain of $g$ are different, so they are not the same function.

doesn't the cancellation of $(x-1)$ allow us to calculate the “true” value of $f(1)$?

I think that your notion of “true value” is exactly what the concept of limit is supposed to encapsulate. As constructed, $f$ is undefined at $1$, but if it were, what should the value be? If we try to use nearby values of $f(x)$ to infer a value for $f(1)$, and if $f(x) = g(x)$ for all $x\neq 1$, then the best answer for $f(1)$ would be $g(1)$.

Matthew Leingang
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