My Math Forum removable discontinuities in rational functions

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 December 4th, 2018, 08:19 PM #1 Senior Member     Joined: Sep 2015 From: USA Posts: 2,463 Thanks: 1340 removable discontinuities in rational functions So I was taught that $f(x) = \dfrac{(x+1)(x-2)}{x+1} = x-2$ is continuous. I'm reading online now that $x=-1$ is considered a removable discontinuity, i.e. a "hole" If this is the case what stops us from creating infinite holes in any continuous function by multiplying by 1 in the fashion $\tilde{f}(x) = f(x)\prod \limits_{k=1}^\infty \dfrac{x-a_k}{x-a_k}$ where $a_k$ is any sequence of real numbers What's the current convention on removable singularities?
 December 5th, 2018, 12:09 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 Nothing stops you from doing that. It's equivalent to not defining the function at infinitely many values of $x$.
December 5th, 2018, 05:08 AM   #3
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 Originally Posted by romsek So I was taught that $f(x) = \dfrac{(x+1)(x-2)}{x+1} = x-2$ is continuous.
Strictly speaking, $\dfrac{(x+1)(x-2)}{x+1} = x-2$ is not true which is essentially the point. The expressions are equal except when $x=-1$ in which case the expression on the left is not defined while the expression on the right is $-3$. In this case, we would define the function $f_1(x) = \dfrac{(x+1)(x-2)}{x+1}$ and the function $f_2(x) = x-2$, and then $f_2$ is referred to as the continuous extension of $f_1$.

This difference is often hammered during a Calc 1 course since students at this level don't understand limits, and they don't understand that functions aren't just formulas.

At the level of a first real analysis course or higher, it is very typical to make no distinction between a function with removable singularities and its continuous extension. For example, a statement like:

"$f(x) = \frac{\sin x}{x}$ is continuous on $\mathbb{R}$" would not be atypical to see and should be considered as a true statement.

What is meant by this, is that $\lim_{x \to 0} f(x) = 1$ so the function being described is assigned the value $f(0) = 1$ which makes it continuous. This is cumbersome to say over and over so when the audience understands limits and continuity a long explanation of this is typically omitted.

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