Question

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & \text{ if }  x < 0 \\ 2x + 3, & x \geq 0\end{cases}\]

Answer 1

When x < 0, then 

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

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

So, the quotient function

\[\frac{\text{ sin } x}{x}\]is continuous at each x < 0. 

When x > 0, then

\[f\left( x \right) = 2x + 3\] , which is a polynomial function.

Therefore,

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

Now,
Let us consider the point x = 0.
Given:

\[f\left( x \right) = \binom{\frac{\text{ sin } x}{x}, \text{ if } x < 0}{2x + 3, \text{ if } x \geq 0 }\]

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) = \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( 2h + 3 \right) = 3\]

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

Thus,

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

Hence, the only point of discontinuity for

\[f\left( x \right)\] is x = 0.