Question

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin x}{x} + \cos x, & \text{ if } x \neq 0 \\ 5 , & \text { if }  x = 0\end{cases}\]

Answer 1

When x  \[\neq\]  0, then

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

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

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

\[\frac{\text{ sin } x}{x}\] Also, cos x is everywhere continuous.
Therefore,

\[\frac{\sin x}{x} + \cos x\]  is continuous at each x \[\neq\] 0.

Let us consider the point x = 0. 

Given: 

\[f\left( x \right) = \binom{\frac{\text{ sin } x}{x} + \ \text{ cos } x, \text{ if }  x \neq 0}{5, \text{ if }  x = 0 }\]

We have
(LHL at x = 0) = \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} \left( \frac{\sin\left( - h \right)}{- h} + \cos\left( - h \right) \right) = \lim_{h \to 0} \left( \frac{\sin\left( - h \right)}{- h} \right) + \lim_{h \to 0} \cos\left( - h \right) = 1 + 1 = 2\]

(RHL at x = 0) = \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( \frac{\sin\left( h \right)}{h} + \cos\left( h \right) \right) = \lim_{h \to 0} \left( \frac{\sin\left( h \right)}{h} \right) + \lim_{h \to 0} \cos\left( h \right) = 1 + 1 = 2\]

\[f\left( 0 \right) = 5\] ∴ \[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) \neq f\left( 0 \right)\]

Thus,

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

Hence, the only point of discontinuity for 

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