For example,

Is the function $h(x)=f(f(x))$ discontinuous at $x=0$ if $f(x)=\frac{1}{x}$ ?

My observation

$h(x)=x$ is continuous at $x=0$

$f(f(x))=\frac{1}{\frac{1}{x}}$ is discontinuous at $x=0$ as $f(x)=\frac{1}{x}$ is discontinuous at $x=0$.

The function $h(x)=x$ and $f(f(x))=\frac{1}{\frac{1}{x}}$ are not exactly the same. Thus, though $h(x)=x$ is continuous, when we say the composite function $f(f(x))=\frac{1}{\frac{1}{x}}=x$ we are already assuming the function is not defined at $x=0$, thus discontinuous at $x=0$.

Is my observation correct ?

Does this applies to similar composite functions, say $h(x)=f(g(x))$ where $f(x)=\frac{1}{x-1}$ and $g(x)=\frac{1}{x-2}$ ?

Similar Post

In a similar problem Discontinuity of composite function , I do not find any explanation rather than few comments arguing the continuity of function $\frac{1}{x}$ does not make sense at $x=0$ and it is a continuous function as the limit exists in its domain. But, check Example 5 clearly states "it is not a continuous function since its domain is not an interval. It has a single point of discontinuity, namely x = 0, and it has an infinite discontinuity there".

  • $\begingroup$ If a function $f$ is not defined at $x_0$, it doesn't have the property of being or not being continuous at that point. You can ask for a continuous extension though. $\endgroup$ – Christoph Apr 10 '18 at 9:44
  • $\begingroup$ @Christoph pls check what is being explained in Example 5 in math.mit.edu/~jspeck/18.01_Fall%202014/Supplementary%20notes/… $\endgroup$ – ss1729 Apr 10 '18 at 9:45
  • $\begingroup$ The definition of continuity of a function in those notes includes the domain being an interval. Note that this a rather non-standard definition. $\endgroup$ – Christoph Apr 10 '18 at 9:48
  • $\begingroup$ h (x) is not equal to x but to $\frac {1}{\frac {1}{x}} $ $\endgroup$ – MysteryGuy Apr 10 '18 at 9:48
  • 1
    $\begingroup$ Well, clearly, $f(x) = \frac1x$ is not continuous at $x = 0$, that we can agree on. However, I subscribe to a notion that it's not discontinuous either. It's just... not defined. There is definitely an asymptote at $x = 0$, though, and in some circumstances I would call it a pole. I guess you could call that an "infinite discontinuity", if you want. $\endgroup$ – Arthur Apr 10 '18 at 11:16

In all your examples, you consider functions that are not defined at some point. If $f$ is not defined at $x=0$, then $g\circ f$ is not defined at $x=0$ either and therefore not continuous at $x=0$.

In more generality, if $f$ is defined over all $\Bbb R$, but not continuous, it may very well be that $g\circ f$ is continuous: this happens for example if $g$ is constant - but of course this is not necessary.


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