Discuss the continuity of the function

Solution:

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.