Question

Find if the given below function is continuous or discontinuous at the indicated point
\(f(x) = \left\{\begin{matrix}
\dfrac{1-\cos 2x}{x^2}, & \text{if}~x \neq 0 \\
5, & \text{if} ~x-0
\end{matrix}\right. ~~\ldots (1)\)

Answer 1

Find right hand limit at \(x = 2\)

Given that,
\(f(x) = \left\{\begin{matrix}
\dfrac{1-\cos 2x}{x^2}, & \text{if}~x \neq 0 \\
5, & \text{if} ~x-0
\end{matrix}\right.~~~\ldots(1)\)
At \(x = 0\)
\(L.H.L = \underset{h \rightarrow 0^-}{\text{lim}} f(x)\)
\(= \underset{h \rightarrow 0^+}{\text{lim}} f(0 - h)\)
\(= \underset{h \rightarrow 0^+}{\text{lim}} \dfrac{1 - \cos2(0 - h)}{(0 -h)^2}\)
\(= \underset{h \rightarrow 0^+}{\text{lim}}\dfrac{2 \sin^2 h}{h^2}~~~~~[\therefore \cos(-x) = cos x]\)
\(= \underset{h \rightarrow 0^+}{\text{lim}}\dfrac{2 \sin^2 h}{h^2}~~~~~[\therefore \cos 2h = 1 - 2 \sin^2 h]\)
\(= 2~~~\ldots(2)~~~~~~~~\left [\because \underset{x \rightarrow 0}{\text{lim}} \dfrac{\sin x}{x} = 1 \right ]\)
Find right hand limit at \(x = 0\)
\(R.H.L = \underset{x \rightarrow 0^+}{\text{lim}} f(x)\)
\(= \underset{h \rightarrow 0^+}{\text{lim}} f(0 + h)\)
\(= \underset{h \rightarrow 0^+}{\text{lim}} \dfrac{1 - \cos 2(0 + h)}{(0 + h)^2}\)
\(= \underset{h \rightarrow 0^+}{\text{lim}}\dfrac{2 \sin^2 h}{h^2} = 2~~~\ldots(3)\)
\(\left\{\because \cos 2x = 1 - 2 \sin^2 x~\text{and}~ \underset{x \rightarrow 0}{\text{lim}}\dfrac{\sin x}{x} = 1 \right\}\)

Also, \(f(0) = 5~~~\ldots(4)\)
Since, \(L.H.L = R.H.L \neq f(0)\)
[From (2), (3), (4)]
\(f(x)\) is discontinuous at \(x = 0\)