# Discontinuity of composite function $f(g(x))$ where $f(x)=\frac{1}{x^2+x-2}$ and $g(x)=\frac{1}{x-1}$

If $f$ and $g$ are both real valued functions such that $f(g(x))$ is defined at $c$. If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then $f(g(x))$ is continuous at $x=c$.

This is fine. But, what if either of the functions is discontinuous ?

What can we say about $f$ and $g$ if $h$ is continuous or discontinuous ?

Ex:1

$f(x)=\frac{1}{x+2}$

In my reference it is said that $f(f(x))$ is discontinuous at $x=-2$ and $x=-5/2$. The observation seems to come from the fact that

$f(x)$ is discontinuous at $x=-2$ and $\frac{1}{x+2}=-2\implies x=-5/2$

But, $f(f(x))=\frac{1}{\frac{1}{x+2}+2}=\frac{x+2}{2x+5}$ seems to be discontinuous only at $x=-5/2$.

Ex:2

$f(x)=\frac{1}{x^2+x-2}$ and $g(x)=\frac{1}{x-1}$

It is said to be discontinuous at $x=1/2, 1, 2$. Pls check link or check Problem, Solution

But, $$f(g(x))=\frac{1}{g^2+g-2}=\frac{1}{(\frac{1}{x-1})^2+(\frac{1}{x-1})-2}=\frac{(x-1)^2}{1+x-1-2x^2+4x-2}\\ =\frac{-(x-1)^2}{2x^2-5x+2}=\frac{-(x-1)^2}{(x-2)(x-\frac{1}{2})}$$ seems to be discontinuous only at $x=2$ and $x=1/2$.

Ex:3

If the observation given in my reference is correct does that mean that in the case where $f(x)=\frac{1}{x}$,

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

Note: Thanx @Andrew Li ,@Math_QED ,@Rob Arthan for their remarks.

My observations are based on the definition that

A real function $f$ is continuous at a point $c$ in the domain if $\lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)$ and

A function is said to be discontinuous (or to have a discontinuity) at some point when it is not continuous there. These points themselves are also addressed as discontinuities, check Real-valued continuous functions . ie, if

1.) $\lim_{x\to c^+}f(x)$ and $\lim_{x\to c^-}f(x)$ exists but are not equal.

2.) $\lim_{x\to c^+}f(x)$ and $\lim_{x\to c^-}f(x)$ exists and are equal, but not equal to $f(a)$.

3.) $f(c)$ is not defined.

• You are wrong. All the functions you consider are continuous on their domain – user370967 Apr 9 '18 at 21:13
• @Math_QED so u mean saying $\frac{1}{x+2}$ is discontinuous at $x=-2$ is wrong ? – ss1729 Apr 9 '18 at 21:15
• Yes. -2 is not in the domain, so it doesn't make sense to talk about continuity at that point – user370967 Apr 9 '18 at 21:16
• @Math_QED So infinite and removable discontinuities don't exist? – Andrew Li Apr 9 '18 at 21:22
• @Math_QED Exactly, and it's has a removable discontinuity. Unless I'm misunderstanding your comment, you say it doesn't make sense to discuss continuity at something not in the domain, so does the removable discontinuity not exist? – Andrew Li Apr 9 '18 at 22:28