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I have been asked to determine whether the function $f(x) = \frac{\sin{x}}{x}$ is continuous on $\mathbb{R}$. I do understand that the function in continuous everywhere except for the point $x=0$, where the function is undefined (the $\frac{0}{0} $ indeterminacy). I do, however, remember what the textbook told me about removable discontinuities, i.e. that whenever a function possesses a definite limit as its argument approaches a value outside of the function’s domain, we are (somehow) given the right to assign the value of that limit to be the value of the function at that point. The question is why in the world such a feint is even plausible? Of course, we can say that the “modified” function with its discontinuity removed is a whole new function, which possesses a domain different from the initial function, that is perfectly fine, but as far as I understand, we regard such functions in their initial form as the ones in which the discontinuities were already removed?

I know that it might be too many words for a question to be even comprehensible, so I am going to provide an example using the function I mentioned already: $f(x) = \frac{\sin{x}}{x}$. We know that it is undefined at $x=0$, however we can remove the discontinuity at that point (for the limit is equal to $1$ as $x$ approaches $0$), and thus make it continuous on $\mathbb{R}$. Do we still regard the modified function as $f(x) = \frac{\sin{x}}{x}$, or do we specifically mention that it is a different function, possessing a distinct value at $x = 0$?

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    $\begingroup$ Without mentioning that it has been redefined, one should not talk of it as being continuous at zero. Although as you say some do say that. $\endgroup$
    – coffeemath
    Commented Sep 26, 2021 at 23:29

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The question is why in the world such a feint is even plausible? $ \newcommand{\set}[1]{\left\{ #1 \right\}} \newcommand{\R}{\mathbb{R}} \newcommand{\f}{\widetilde{f}} $

Good question. It's more of a sleight of hand than "something we can do."

So we're on the same page, let $f : D \setminus \set{a} \to \R$ be such that $$ \lim_{x \to a} f(x) = b $$ This is ultimately the situation we're always dealing with when contending with removable discontinuities: the function is not defined at a point (hence why we write $D \setminus \set{a}$, where $D$ will be the domain when $a$ has a defined output), but the limit as we get there still exists.

When we say "we can write $f(a) = b$ to make $f$ defined and continuous at $a$", what we're really doing is defining a new function $\f$ which is a continuous extension of $f$. Specifically $$ \f : D \to \R \text{ and is defined by } \f(x) = \begin{cases} f(x) & x \in D \setminus \set{a} \\ \displaystyle \lim_{x \to a} f(x) = b & x = a \end{cases} $$ From here, typically in a lower-level calculus class, we would just give $\f$ the same name as $f$, which can result in confusion. After all, $\f$ and $f$ are not the same function: $\f(a)$ exists, whereas $f(a)$ does not. (Even though the limit does exist, $f(a)$ still does not exist.)

Leaving the name the same is the "sleight of hand" I referred to, or perhaps considered an abuse of notation. Ultimately, calculus students usually don't need to know that technically $\f$ and $f$ are not the same functions because, in calculus, their goal is usually to just remove that discontinuity in the first place. They don't have the formal knowledge necessary to properly understand why this is somewhat problematic, since, at that level, it does make some sort of hand-wavey sense.

Though as a calculus instructor now myself, I'm not a fan of it, and am a bit more careful to articulate what's going on, or to at least word things like "if we were to redefine $f$ so it is defined at $a$, what value of $f(a)$ would make it continuous?"

$f(x) = \frac{\sin{x}}{x}$. We know that it is undefined at $x=0$, however we can remove the discontinuity at that point (for the limit is equal to 1 as x approaches 0), and thus make it continuous on $\mathbb{R}$. Do we still regard the modified function as $f(x) = \frac{\sin{x}}{x}$, or do we specifically mention that it is a different function, possessing a distinct value at $x = 0$?

So, in short -- formally, we recognize it a different function. Some lower-level textbooks might just use the same name for it, creating this cognitive dissonance in the astute observers, but ultimately the function with the removed discontinuity is indeed a different function (even though it is a continuous extension and may have many of the same properties).

We would write

$$\f : \R \to \R \text{ is defined by } \f(x) = \begin{cases} f(x) = \sin(x)/x & x \ne 0 \\ \displaystyle \lim_{x \to 0} f(x) = 1 & x = 0 \end{cases}$$

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  • $\begingroup$ Kind of funny is the fact that the textbook I use does not denote the re-defined function as the initial one, nor does it mention that it is a whole different one. Nevertheless, thank you for reaching out so quickly! Much obliged. $\endgroup$
    – Barbatulka
    Commented Sep 26, 2021 at 23:39

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