# Justification for cancellation in rational functions

1. Feb 18, 2015

### Mr Davis 97

For example, say we have $\frac{x^4(x - 1)}{x^2}$. 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 $x^2(x - 1)$, then $x^2(x - 1)(1)$, and since $\frac{x^2}{x^2} = 1$, we then have $\frac{x^4(x - 1)}{x^2}$. 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 $\frac{x^4(x - 1)}{x^2}$, how to we know that cancelling the x's will lead to the correct limit, since that is in effect the limit of the function $x^2(x−1)$ and not $\frac{x^4(x - 1)}{x^2}$?

2. Feb 19, 2015

### axmls

We can't in general do that and say they're the same function. However, there's a property of limits that, given a function $f$ and a function $g$ such that $g(x) = f(x)$ except maybe at a point $x = c$, then $$\lim_{x \to c} f(x) = \lim_{x \to c} g(x)$$ What we care about is the limit. We're not worried about what actually happens at the point. We're only worried about what happens as we approach the point. What's important here is that we know that the two functions are equal everywhere except that one point.

I may be able to find a specific statement of this.

3. Feb 19, 2015

### axmls

The following problem comes from Chapter 5, problem 10 of Spivak's Calculus 1st edition:

If you're familiar with epsilon-delta proofs (the limit approaches $L$ if $0 < |x - a| < \delta$ implies $|f(x) - L| < \epsilon$), then this is easy to see: We know that $f(x) = g(x)$ for some neighborhood around $a$, that is, $f(x) = g(x)$ when $|x - a| < \delta '$. Furthermore, if $\lim_{x \to a} f(x) = L$ then given an $\epsilon > 0$, then there exists some $\delta > 0$ such that $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$. Now just make sure $\delta ' < \delta$, and then we know $g(x)$ is equivalent to $f(x)$, so we can replace it in our mathematical statement: $0 < |x - a| < \delta ' \implies |g(x) - L| < \epsilon$. Thus $f$ and $g$ both approach $L$ as $x \to a$.

Intuitively, this just means that we only care about the neighborhoods near the point we're approaching. Since the two functions are equal everywhere except that one point, the two limits are still the same.