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Why are we allowed to cancel the x's in a expression such as $\dfrac{x^2(x-1)}{x^4}$? Don't we lose information about the function when we do, such as the fact that the function is undefined at $x = 0$? Specifically, concerning limits, why are we allowed to cancel variables in the rational expression and evaluate the limit as if it applied to the original function? How do we know for sure that each time the function that results from cancellation will yield the correct limit as it applies to the function before cancellation?

Mihail
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Wesley
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  • In the particular example you chose, the fraction after simplification will still have an $x$ in the denominator, so the domain is clear. Perhaps a better example for your question is $\dfrac xx$. – GFauxPas Feb 18 '15 at 22:43

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The reason we don't care about the domain of a function at the limit point is because it's wholly irrelevant to the limit.

(Let me know if the following notation is confusing, I can change the answer to use more natural language if you wish.)

The statement:

$$f(x) \to \ell \text{ as } x \to c$$

means:

$$\forall \epsilon > 0 : \exists \delta > 0:0 < \vert x - c \vert < \delta \implies \vert f(x) - \ell \vert < \epsilon $$

In particular, let's focus on:

$$0 < \vert x - c \vert$$

This part of the definition forces $x \ne c$. So if after simplification something happens differently at $x = c$, it is nothing to be concerned about. Nothing in the definition of limit is affected at all by what happens at $x = c$.

Formal exposition of what I said: If two functions agree in a punctured neighborhood of the limit point (in the case above, the punctured neighborhood is some real interval containing $c$ but with $c$ removed), then the limit, if it exists, will be the same for both functions.

GFauxPas
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