Continuity of Composite Function - Practice Questions & MCQ

Continuity of Composite Function 

If the function $f(x)$ is continuous at the point $x=a$ and the function $y=g(x)$ is continuous at the point $x=f(a)$, then the composite function $y=(g \circ f)(x)=g(f(x))$ is continuous at the point $x=a$.

Consider the function $f(x)=\frac{1}{1-x}$, which is discontinuous at $x=1$ If $g(x)=f(f(x))$
$g(x)$ will not be defined when $f(x)$ is not defined, so $g(x)$ is discontinuous at $x=1$
Also $g(x)=f(f(x))$ is discontinuous when $f(x)=1$
i.e. $\frac{1}{1-x}=1 \Rightarrow x=0$

We ca check it by finding $g(x), \quad g(x)=\frac{1}{1-f(x)}=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$
It is discontinuous at $\mathrm{x}=0$
So, $g(x)=f(f(x))$ is discontinuous at $x=0$ and $x=1$

Now consider,

$
\begin{aligned}
\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x}))) & =\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right) \\
& =\frac{1}{1-\frac{\mathrm{x}-1}{\mathrm{x}}}=\mathrm{x}
\end{aligned}
$

seems to be continuous, but it is discontinuous at $x=1$ and $x=0$ where $f(x)$ and $f(f(x))$ respectively are not defined.