 # Prove that

Question:

Prove that

$f(x)= \begin{cases}\frac{\sin x}{x}, & x<0 \\ x+1, & x \geq 0\end{cases}$

is everywhere continuous.

Solution:

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.