Find the points of discontinuity of f, where
Question:

Find the points of discontinuity of f, where

$f(x)=\left\{\begin{array}{l}\frac{\sin x}{x}, \text { if } x<0 \\ x+1, \text { if } x \geq 0\end{array}\right.$

Solution:

The given function $f$ is $f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \\ x+1, & \text { if } x \geq 0\end{cases}$

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

If $c<0$, then $f(c)=\frac{\sin c}{c}$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(\frac{\sin x}{x}\right)=\frac{\sin c}{c}$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $f(c)=c+1$ and $\lim f(x)=\lim (x+1)=c+1$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Thereforef is continuous at all points x, such that x > 0

Case III:

If $c=0$, then $f(c)=f(0)=0+1=1$

The left hand limit of f at x = 0 is,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

The right hand limit of f at x = 0 is,

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(x+1)=1$

$\therefore \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

Thereforef is continuous at x = 0

From the above observations, it can be concluded that f is continuous at all points of the real line.

Thus, f has no point of discontinuity.