Mr Davis 97

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 $\displaystyle \frac{x^4(x - 1)}{x^2}$?

v8archie

Strictly speaking, you can't without explicitly excluding 0 from the domain. 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 $$\lim_{x \to 0} {x^4(x-1) \over x^2} = \lim_{x \to 0} \left({x^2 \over x^2}x^2(x-1)\right) = \left(\lim_{x \to 0} {x^2 \over x^2} \right)\bigg(\lim_{x \to 0} x^2(x-1)\bigg) = 1 \cdot 0 = 0$$

skipjack

As x approaches zero, it is, by definition, non-zero, so cancelling it is allowed.

CRGreathouse

This example may help.

Let $f(x)=x$ when $x\ne3$ and $f(3)=\text{hat}.$ What is $\lim_{x\to3}f(x)$?