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I'll state a question from my textbook below:

Given $f(x) = \frac 1 {x-1}$. Find the points of discontinuity of the composite function $y = f[f(x)]$.

Clearly, $f(x)$ is not defined at $x=1$. But that is not the case with $y = \frac {x-1}{2-x}$. Calculating the limit and the value of $y$ at $x=1$ we even find that it is continuous at this point. The graph of $y$ gives the same idea.

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But my textbook says that it is discontinuous at $x=1$. I know this is because the continuity of the composite function $f\circ g$ at $c$ requires the continuity of $g$ at $c$. That is exactly what I can't understand. Why? I don't understand what does $f\circ g$ have to do with the continuity of $g$ at any point, say $c$? It only needs $g$ to be defined at $c$, right? Do you have an example that illustrates the importance of continuity of a function $g$ at $c$ for the continuity of the composite function $f\circ g$ at $c$?

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  • $\begingroup$ In order to find the value of the composite function $f \circ f$ at $1$, you first calculate $f(1)$ and then $f(f(1))$. The first value is not defined, so the composite function is not defined. $\endgroup$ – Eivind Apr 17 '18 at 13:54
  • $\begingroup$ A bad habit of your textbook: as $f$ isn't defined in $x=1$ it can't be neither continuous nor discontinuous. $\endgroup$ – Michael Hoppe Apr 17 '18 at 14:02
  • $\begingroup$ @MichaelHoppe I thought a function is always discontinuous at a point where it is not defined. $\endgroup$ – SamInuyasha ANMF Apr 17 '18 at 15:29
  • $\begingroup$ @SamInuyashaANMF Of course not, see math.stackexchange.com/questions/1482787/… $\endgroup$ – Michael Hoppe Apr 17 '18 at 21:53
  • $\begingroup$ @MichaelHoppe I thought there's a difference between $f(x) = \frac 1x$ and $g(x) = \frac 1x, x \ne 0$. $0$ is in the domain of $f$, whereas it is not in the domain of $g$. So, $f$ is discontinuous at $x = 0$, whereas there's no point in discussing continuity of $g$ at $x = 0$ since the point is not even in its domain. $\endgroup$ – SamInuyasha ANMF Apr 18 '18 at 6:59
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The answer is that the function $$ f(f(x)) = \frac{1}{\frac{1}{x - 1} - 1} $$ is not the same as $$ \frac{x - 1}{2 - x}. $$ They give the same outputs for all $x$ in the domain of $f(f(x))$, but the second function has $1$ in its domain. The reason is that the limit of $f(f(x))$ exists at $1$, but it does not have a value. The second function just fills in the value with the limit. When you manipulate $f(f(x))$ to reduce it to the second function, you probably multiplied by $x - 1$ or something. You were making the implicit assumption that $x - 1 \neq 0$. But, it could be. So, you changed the function at that point in your work.

Then $f(f(x))$ is not continuous at $x = 1$, because there is a division by $0$ error when you attempt to calculate it. But, in order to be continuous it needs both the limit and no computation error.

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    $\begingroup$ It doesn't only need those limits, but a defined value of $f(1)$ as well. Continuity at $x= isn't a question at all. $\endgroup$ – Michael Hoppe Apr 17 '18 at 14:06
  • $\begingroup$ Supposing a random function $g$ defined, but not continuous, at $c$, will $f \circ g$ be continous at $c$ for another random function $f$ [which is continuous at $g(c)$], such that $f[g(c)] = \lim \limits_{x \to g(c)} f(x)$? $\endgroup$ – SamInuyasha ANMF Apr 17 '18 at 15:26
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Technically, if $f(x) = \frac{1}{x-1}$, then $$f\circ f(x) = \frac{1}{\left(\frac{1}{x-1}\right)-1} = \frac{1}{\left(\frac{2-x}{x-1}\right)}$$ This is not the same as $$g(x) = \frac{x-1}{2-x}$$ They are the same almost everywhere, where the only point they differ is $x=1$ ($f\circ f$ is not defined there whereas $g$ is). In general, a fraction $1/(a/b) = b/a$ if you can assume $b\not=0$. Otherwise, such an operation is not defined.

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    $\begingroup$ This is a good answer, and teaches a general rule at the same time. I will use this in the fall when I teach Calculus. $+1$ $\endgroup$ – Joe Johnson 126 Apr 17 '18 at 14:01
  • $\begingroup$ Supposing a random function $g$ defined, but not continuous, at $c$, will $f \circ g$ be continous at $c$ for another random function $f$ [which is continuous at $g(c)$], such that $f[g(c)] = \lim \limits_{x \to g(c)} f(x)$? $\endgroup$ – SamInuyasha ANMF Apr 17 '18 at 15:26
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    $\begingroup$ @SamInuyashaANMF I don't know what you mean about "random" but here is a counter example I think that works. Let $g(x)$ be the Dirchlet function, that is $g(x) =1$ if $x$ is rational and $=0$ if $x$ is irrational. Obviously $g$ is not continuous anywhere. Then let $f(x) = x$. Clearly this is continuous at any $g(c)$ but $f\circ g$ is certainly not continuous. $\endgroup$ – welshman500 Apr 17 '18 at 16:40
  • $\begingroup$ @welshman500 I get your point. Suppose $g(x) = x$, if $x \ne 1$ and $g(x) = 0$, if $x = 1$. Let $f(x) = \frac {x-1}{x-1}$. Here, $g$ is discontinuous at $x = 1$ but $f \circ g$ is continuous. So continuity of $g$ at $c$ is not always necessary, right? But yes, you gave the right example to show the importance of the continuity of $g$. Thanks! $\endgroup$ – SamInuyasha ANMF Apr 18 '18 at 6:54

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