How $f\circ g$ is discontinuous at $x=2$?

Question

Given $f(x)=\dfrac{1}{x-1}$ and $g(x)=\dfrac{1}{x-2}$, it is obvious that both are having infinite discontinuities at $x=1$ and $x=2$. Now for$$(f \circ g)(x)=\frac{x-2}{3-x},$$ where is it discontinuous? No doubt at $x=3$, but why $(f \circ g)(x)$ is also discontinuous at $x=2$?

Answer 1

You have $f(x)=\frac{1}{x-1}$ and $g(x)=\frac{1}{x-2}$ ; so

$f(g(x))=\frac{1}{g(x)-1} = \frac{1}{\frac{1}{x-2}-1}=\frac{x-2}{3-x}$ ;

Consider the second equation for your answer If (g(x)) would not exist at all at (x=2) then where would you find (f(g(x)) to compute?

Answer 2

To answer your second question: $f \circ g$ means that you first apply $g$ and then apply $f.$ You cannot apply $g$ to $x=2$, so $f \circ g$ is not defined at $x=2.$

Answer 3

By definition, ($f$ $\circ$ $g$)($x$) = $f(g(x))$. So ($f$ $\circ$ $g$)$(2)$ = $f(g(2))$. But $g$ is discontinuous at $x=2$, so $f \circ g$ is not defined for $x=2$, so it's not continuous.