The points of discontinuity of the function $f(x)=\frac{1}{x-1}$ if $0 \leq x \leq 2$ $=\frac{x+5}{x+3}…
The points of discontinuity of the function $f(x)=\frac{1}{x-1}$ if $0 \leq x \leq 2$ $=\frac{x+5}{x+3} \quad$ if $\quad 2 < x \leq 4$ in its domain are
x=2 only
x=1, x=2
x=4 only
x=0, x=2
Solution
$\begin{aligned}
f(x) &=\frac{1}{x-1}, \text { if } 0 \leq x \leq 2 \Rightarrow f(x) \text { is not defined at } x=1 \\
&=\frac{x+5}{x+3}, \text { if } 2 < x \leq 4
\end{aligned}$
$\lim _{x \rightarrow 2^{-}} f(x)=\frac{1}{2-1}=1 \text { and } \lim _{x \rightarrow 2^{+}} f(x)=\frac{7}{5}$
Thus $f(x)$ is not continuous at $x=2$