Question

Prove that the function \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & x < 0 \\ x + 1, & x \geq 0\end{cases}\]  is everywhere continuous.

 

Answer 1

When x < 0, we have  

\[f\left( x \right) = \frac{\text{ sin } x}{x}\]

We know that sin x as well as the identity function x are everywhere continuous. So, the quotient function 

\[\frac{\text{ sin } x}{x}\]

When x > 0, we have

\[f\left( x \right) = x + 1\]

Therefore,

\[f\left( x \right)\] is continuous at each x > 0.

Given:

\[f\left( x \right) = \binom{\frac{\text{ sin } x}{x}, x < 0}{x + 1, x \geq 0}\]

We have
(LHL at x = 0) = \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} \left( \frac{\sin\left( - h \right)}{- h} \right) = \lim_{h \to 0} \left( \frac{\sin\left( h \right)}{h} \right) = 1\]

(RHL at x = 0) =

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( h + 1 \right) = 1\]

Also, 

\[f\left( 0 \right) = 0 + 1 = 1\]

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]

Thus

\[f\left( x \right)\]  is continuous at x = 0.

Hence,

\[f\left( x \right)\] is everywhere continuous.