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Is there any value, say, to $g(3)$ where $g(x) := (x-3)*\dfrac{x}{(x-3)}$ ?

I read something where it seems as though they assumed it is $0$ (for an expression of this sort). Is it true? If so, why? If $\dfrac{3}{(3-3)}$ is not a real number, or any number- why are we motivated to define $0$ times 'it' to be $0$ (if that indeed is the case)?

Eliyahu Abadi
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    Maybe you misread, or maybe the source is plain wrong. $g(3)$ is not defined as written. It could be defined by continuity to be $g(3)=3$, but defining it as $0$ makes no sense (unless there is a great deal of other context missing from the question). – dxiv Jun 10 '21 at 18:39
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    Totally agreeing with @dxiv's point, it would be better if you tell the source of the problem, and read the context again to see if you're missing something. – ultralegend5385 Jun 10 '21 at 18:41
  • If $g(x)=\frac{x(x-3)}{x-3}$, then $g(3)$ is not defined because division by zero is always not defined. However, $$ \lim_{x \to 3}g(x)=3 , . $$ In other words, the function $g$ approaches $3$ near $3$ because we can make $g(x)$ as close to $3$ as we like by requiring that $x$ be close to, but unequal to, $3$. If we define a new function $\hat{g}$ such that $\hat{g}(x)=x$, then $\lim_{x \to 3}g(x)=g(3)$, meaning that $g$ has a continuous extension at $3$. – Joe Jun 10 '21 at 18:58

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