Prove that the function


Prove that the function $f(x)=\left\{\begin{array}{ll}\frac{\sin x}{x}, & x<0 \\ x+1, & x \geq 0\end{array}\right.$ 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 $x<0$.


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

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left(\frac{\sin (-h)}{-h}\right)=\lim _{h \rightarrow 0}\left(\frac{\sin (h)}{h}\right)=1$

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



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

Thus, $f(x)$ is continuous at $x=0$.

Hence, $f(x)$ is everywhere continuous.

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