Are polynomial fractions and their reductions really equal?

Question

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).

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 $x=1$.

Answer 1

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 $x = 1$.

Answer 2

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 $g(1)$.