Question

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & \text{ if }   x \neq 0 \\ 4 , & \text{ if }  x = 0\end{cases}\]

 

Answer 1

When x \[\neq\]  0, then

\[f\left( x \right) = \frac{\sin3x}{x}\]

We know that sin 3x as well as the identity function x are everywhere continuous. So, the quotient function

\[\frac{\sin3x}{x}\]  is continuous at each x \[\neq\] 0. 

Let us consider the point x = 0. 

Given: 

\[f\left( x \right) = \binom{\frac{\sin3x}{x}, \text{ if }  x \neq 0}{4, \text{ if } x = 0 }\]

We have

(LHL at x = 0) = \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( \frac{\sin \left( 3h \right)}{h} \right) = \lim_{h \to 0} \left( \frac{3 \sin \left( 3h \right)}{3h} \right) = 3\]

(RHL at x = 0) =\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( \frac{\sin \left( 3h \right)}{h} \right) = \lim_{h \to 0} \left( \frac{3 \sin \left( 3h \right)}{3h} \right) = 3\]

Also,

\[f\left( 0 \right) = 4\]

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

Thus,

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

Hence, the only point of discontinuity for

\[f\left( x \right)\] is x = 0.