Question

Find all points of discontinuity of $f$, where
f(x) = $\{\begin{array}{c}\frac{\sin x}{x},\text{if}x<0\\ x+1,\text{if}x\ge 0\end{array}$

Answer 1

The given function is $f\left(x\right)=\{\begin{array}{c}\frac{\sin x}{x},\text{if}x<0\\ x+1,\text{if}x\ge 0\end{array}$
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 $f\left(c\right)=\frac{\sin c}{c}$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}\frac{\sin x}{x}=\frac{\sin c}{c}$
Therefore, $f$ is continuous at all points $x$, such that $x<0$
Case II
If $c>0$, then $f\left(c\right)=c+1$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x+1)=c+1$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points such that $x>0$
Case III
If $c=0$, then $f\left(c\right)=f\left(0\right)=0+1=1$
The left hand limit of $f$ at $x=0$ is,
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\sin x}{x}=1$
The right hand limit of $f$ at $x=0$ is,
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}(x+1)=1$
$\therefore \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)=f\left(0\right)$
Therefore, $f$ 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.