Discuss the continuity of the function
$f(x)= \begin{cases}2 x-1, & \text { if } x<2 \\ \frac{3 x}{2}, & \text { if } x \geq 2\end{cases}$
When $x<2$, we have
$f(x)=2 x-1$
We know that a polynomial function is everywhere continuous.
So, $f(x)$ is continuous for each $x<2$.
When $x>2$, we have
$f(x)=\frac{3 x}{2}$
The functions $3 x$ and 2 are continuous being the polynomial and constant function, respectively. Thus, the quotient function $\frac{3 x}{2}$ is continuous at each $x>2$.
Now,
Let us consider the point $x=2$.
Given: $f(x)=\left\{\begin{array}{l}2 x-1, \text { if } x<2 \\ \frac{3 x}{2}, \text { if } x \geq 2\end{array}\right.$
We have
$(\mathrm{LHL}$ at $x=2)=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}(2(2-h)-1)=4-1=3$
$(\mathrm{RHL}$ at $x=2)=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} \frac{3(h+2)}{2}=3$
Also,
$f(2)=\frac{3(2)}{2}=3$
$\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$
Thus, $f(x)$ is continuous at $x=2$.
Hence, $f(x)$ is everywhere continuous.