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December 4th, 2018, 09:19 PM  #1 
Senior Member Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157  removable discontinuities in rational functions
So I was taught that $f(x) = \dfrac{(x+1)(x2)}{x+1} = x2$ 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 $\tilde{f}(x) = f(x)\prod \limits_{k=1}^\infty \dfrac{xa_k}{xa_k}$ where $a_k$ is any sequence of real numbers What's the current convention on removable singularities? 
December 5th, 2018, 01:09 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,986 Thanks: 1853 
Nothing stops you from doing that. It's equivalent to not defining the function at infinitely many values of $x$.

December 5th, 2018, 06:08 AM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 521 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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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discontinuities, functions, rational, removable 
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