Prove that
$f(x)= \begin{cases}\frac{\sin x}{x}, & x<0 \\ x+1, & x \geq 0\end{cases}$
is everywhere continuous.
When $x<0$, we have
$f(x)=\frac{\sin x}{x}$
We know that $\sin x$ as well as the identity function $x$ are everywhere continuous.
So, the quotient function $\frac{\sin x}{x}$ is continuous at each $x<0$.
When $x>0$, we have
$f(x)=x+1$, which is a polynomial function.
Therefore, $f(x)$ is continuous at each $x>0$.
Now,
Let us consider the point $x=0$.
Given: $f(x)=\left\{\begin{array}{l}\frac{\sin x}{x}, x<0 \\ x+1, x \geq 0\end{array}\right.$
We have
(LHL at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left(\frac{\sin (-h)}{-h}\right)=\lim _{h \rightarrow 0}\left(\frac{\sin (h)}{h}\right)=1$
$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(h+1)=1$
Also,
$f(0)=0+1=1$
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
Thus, $f(x)$ is continuous at $x=0$.
Hence, $f(x)$ is everywhere continuous.