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I was reading Thomas' calculus which said that $y=1/x$ is a continuous function because it was continuous at every point of its domain(it not being defined at $x=0$), but then goes on to show that $1/x^2$ is discontinuous saying it is an infinity discontinuity as $x$ approaches $0$. Why doesn't the same logic apply to $y=1/x$?

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    $\begingroup$ I haven't read that particular book, but both $\frac{1}{x}$ and $\frac{1}{x^2}$ are continuous on $\mathbb{R}\setminus\{0\}$. $\endgroup$ – Aweygan Sep 13 '16 at 4:50
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    $\begingroup$ I don't have a copy of Thomas at hand, but I would suspect that it's more likely that you are misinterpreting Thomas than that he made such a blunder. $\endgroup$ – Robert Israel Sep 13 '16 at 5:15
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    $\begingroup$ Observe that $\frac{1}{x}\rightarrow +\infty$ as $x\rightarrow 0+$ and $\frac{1}{x}\rightarrow -\infty$ as $x\rightarrow 0-$. On the other hand, $\frac{1}{x^2}\rightarrow +\infty$ from both sides. He may be talking about that. But I'm not sure you interpreted the whole thing correctly. $\endgroup$ – JDF Sep 13 '16 at 5:40
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    $\begingroup$ @DerbyMoose are you able to post some images of the page(s) for context? (With low rep you can't post your's here directly, but put them on Imgur or something and that will suffice) $\endgroup$ – Brevan Ellefsen Apr 29 '17 at 19:41
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    $\begingroup$ You could extend $1/x^2$ to a continuous function from the real line into its two-point compactification $[-\infty,\infty]$, simply by mapping $0$ to $\infty$. This is not possible with $1/x$, though you could extend that one to a continuous function into the one-point compactification of $\mathbb{R}$. I haven't read Thomas' calculus text, but this is the only way I can think of to justify the statement. $\endgroup$ – Harald Hanche-Olsen Jun 8 '17 at 7:57
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The notion of continuity at $x_0$ only applies to points of the domain.
By definition, a function is said continuous if it is continuous at each point of its domain.

The notion of discontinuity does not apply only to points of the domain but also to limit point of the domain that do not belong to the domain. So a continuous function can have discontinuities, but not in points of the domain.

As an example, the function $$ f(x)=\begin{cases} -1, & x<0 \\ +1, & x\geq0 \end{cases} $$ is not continuous because the limit in $0$ does not exist.
The function $$ f(x)=\begin{cases} -1, & x<0 \\ +1, & x>0 \end{cases} $$ is continuous, because it is continuous in each point of its domain (note that in this case $0$ does not belong to the domain). This function has however a discontinuity in $0$.

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$\frac{1}{x}$ is continuous in the domain. As zero is not in the domain, you can say $\frac{1}{x}$ is a continuous function. In Thomas' book, the example of $\frac{1}{x^2}$ is not continuous on a point, which is zero. If zero, which is not in domain of $\frac{1}{x^2}$, is exclude, you can say $\frac{1}{x^2}$ is a continuous function. If you are referring to $\mathbb R$, the natural number, then $\frac{1}{x^2}$ is not continuous.

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Short answer: The same logic does apply for $y = \frac{1}{x}$.

Long answer: A continuous function is defined to have no discontinuities within its domain. Therefore, $y = \frac{1}{x}$ is a continuous function because $x = 0$ is not part of its domain. However, there is an infinite discontinuity at $x = 0$, but because $x = 0$ is not part of the function's domain, $y = \frac{1}{x}$ is still considered to be a continuous function. Using the same logic, $y = \frac{1}{x^2}$ is also a continuous function, just with a discontinuity at $x = 0$, which is not part of its domain.

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In order for an equation, f(x), to be continuous is that for each point on the interval of it's domain, the value must exist. However, should certain values be excluded from the function's domain, then it is not necessary that they exist, and the function will remain continuous. Additionally, if the limit of f(x) for a certain portion that is excluded from the domain does exist, and does not contain a physical point in that region, then the graph will be deemed discontinuous. For example, in the graph (1/x), on the interval of (-∞,∞) the domain includes all real numbers except X=0. However, since the domain excludes x=0, and the limit of this function at 0 does not exist, then it is not necessary that this be a point on the graph. For this reason, this graph is continuous. Additionally, in the graph (1/x^2), on the interval of (-∞,∞) the domain includes all real numbers except X=0. However, since the domain excludes x=0, but the limit of this function at 0 does actually exist (answer is ∞), then it is necessary that this be a point on the graph. Considering the graph does not contain a point at this location, it is understood to be discontinuous. This is my interpretation :)

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  • $\begingroup$ Welcome to math.SE. You could use MathJax to format your statement to increase readability. Plus not a good idea to use graph to prove discontinuity - you could draw a graph for illustration/intuition. $\endgroup$ – Yujie Zha Jun 30 '17 at 1:13

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