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Let $f(x) = \frac{x}{x}$.

Is it correct to say that $f(x) \ne 1$, since $f(x)$ has a discontinuity at $x=0$?

Mikasa
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Kenshin
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    $f$ does not have a discontinuity at $0$, it is simply not defined there. – Tobias Kildetoft Oct 18 '12 at 13:28
  • Isn't it discontinuous because the limit as f(x) approaches 0 is not equal to f(0)? – Kenshin Oct 18 '12 at 13:30
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    No, the value $f(0)$ is not defined, so you cannot compare anything to it. – Tobias Kildetoft Oct 18 '12 at 13:30
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    @Tobias If both one-sided limits exist and are equal, but the function is not defined there, it is often called a "removable discontinuity". – Austin Mohr Oct 18 '12 at 13:31
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    There's a distinction between "not continuous" and "discontinuous".. – The Chaz 2.0 Oct 18 '12 at 13:34
  • Ok thanks guys these comments have been useful. – Kenshin Oct 18 '12 at 13:34
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    I agree with Austin Mohr. It may depend on what level you're talking about, but in undergraduate calculus, which is probably all that matters here, this is called a removable discontinuity. The three part definition for continuity given in most calculus books is 1. The limit exists, 2. The function is defined at that value, 3. The two values are equal to each other. And, when one of those fails, then $f$ is discontinuous there. And, by the way, no distinction is ever created about "not continuous" or "discontinuous". In the book I am looking at, both are used to mean the same thing. – GeoffDS Oct 18 '12 at 14:55

3 Answers3

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It depends on the domain of $f$ and $1$. If these are $\Bbb R\setminus\{0\}$ and $\Bbb R$, respectively, from a formal perspective the functions will not be equal. Restricted to a domain $S \subseteq \Bbb R$ where both are well-defined (i.e. not containing $0$), they are equal.

Lord_Farin
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Exactly, $f$ is the constant $1$ function on its domain $\Bbb R\setminus\{0\}$.

Berci
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A given function always comes with a domain. For two functions to be equal (in a formal sense) the domains need to be the same.

For example consider the functions: $$ f: \mathbb{R} \to [0,\infty) $$ given by $$ f(x) = x^2 $$ and the function $$ g: [0,\infty) \to [0,\infty) $$ given by $$ g(x) = x^2. $$ These two functions are not the same even though they are given by the same expression. Note for example that $g$ is one-to-one (injective), while $f$ is not. (If the functions were the same, then one would think that they should have exactly the same properties.)

Now, often we don't specify the domain. We just talk about the function defined by (say) $h(x) = \frac{1}{x}$. In such a case we usually assume that the domain is the set of all $x$ ]for which the expression makes sense. Hence the domain of $h$ is $\mathbb{R}\setminus \{0\}$.

Now then, when you write $f(x) = \frac{x}{x}$, then I would say that the domain is $\mathbb{R}\setminus\{0\}$, because I cannot evaluate the expression at $0$. The domain of the function $g(x) = 1$ is all real numbers. So we would not say that the functions are equal.

Note also that to cancel $x$ in the expression $\frac{x}{x}$, $x$ would have to be non-zero.

Edit: About the discontinuity argument, I notice that in the comments to your questions people are saying that "discontinuous" doesn't mean "not continuous". I am not sure that this is a universally agreed upon notion. According to Wikipedia: "If a function is not continuous at a point in its domain, one says that it has a discontinuity there." So Wikipedia seems to agree with the comments. According to Wolfram MathWorld: discontinuous means not continuous. In any case, you are right in noting that the functions are not equal because the function $f(x) = \frac{x}{x}$ is not continuous at $0$.

Thomas
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