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 February 18th, 2015, 07:48 PM #1 Senior Member     Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus How to justify the cancelling of variables in a rational expression? For example, say we have $\displaystyle \frac{x^4(x - 1)}{x^2}$. The function is undefined at 0, but if we cancel the x's, we get a new function that is defined at 0. So, in this case, we have $\displaystyle x^2(x - 1)$, then $\displaystyle x^2(x - 1)(1)$, and since $\displaystyle \frac{x^2}{x^2} = 1$, we then have $\displaystyle \frac{x^4(x - 1)}{x^2}$. However, this is a new function, since the domain has changed to exclude x = 0. How is this justified? Why can we go about changing the function in that way? Specifically, when we evaluate limits, in the case where we have $\displaystyle \frac{x^4(x - 1)}{x^2}$, how do we know that cancelling the x's will lead to the correct limit, since that is in effect the limit of the function $\displaystyle x^2(x - 1)$ and not $\displaystyle \frac{x^4(x - 1)}{x^2}$? Last edited by skipjack; February 18th, 2015 at 11:25 PM.
February 18th, 2015, 08:59 PM   #2
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Quote:
 Originally Posted by Mr Davis 97 How is this justified? Why can we go about changing the function in that way?
Strictly speaking, you can't without explicitly excluding 0 from the domain.
Quote:
 Originally Posted by Mr Davis 97 Specifically, when we evaluate limits... how do we know that cancelling the x's will lead to the correct limit?
This is the exception to the rule, it's valid because the limit is not the function. The limit tells us where the function is heading towards, but is silent on whether the function actually reaches that point.

${x^2 \over x^2} = 1$ for all $x \ne 0$ so it is clear that it is heading towards 1 for $x=0$ too - this can be shown analytically too. We also have the formula $$\left.\begin{array}{c}\lim_{x \to a} f(x) = F \\ \lim_{x \to a} g(x) = G\end{array}\right\} \implies \lim_{x \to a} f(x)g(x) = FG$$
And thus $$\lim_{x \to 0} {x^4(x-1) \over x^2} = \lim_{x \to 0} \left({x^2 \over x^2}x^2(x-1)\right) = \left(\lim_{x \to 0} {x^2 \over x^2} \right)\bigg(\lim_{x \to 0} x^2(x-1)\bigg) = 1 \cdot 0 = 0$$

Last edited by skipjack; February 18th, 2015 at 11:26 PM.

February 18th, 2015, 11:22 PM   #3
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Quote:
 Originally Posted by Mr Davis 97 Specifically, in the case where we have $\displaystyle \frac{x^4(x - 1)}{x^2}$, how do we know that cancelling the x's will lead to the correct limit?
As x approaches zero, it is, by definition, non-zero, so cancelling it is allowed.

 February 19th, 2015, 06:16 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms This example may help. Let $f(x)=x$ when $x\ne3$ and $f(3)=\text{hat}.$ What is $\lim_{x\to3}f(x)$?

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