Question

If the functions f(x), defined below is continuous at x = 0, find the value of k. \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2x}{2 x^2}, & x < 0 \\ k , & x = 0 \\ \frac{x}{\left| x \right|} , & x > 0\end{cases}\] 

 

Answer 1

Given: 

\[f\left( x \right) = \begin{cases}\frac{1 - \cos2x}{2 x^2}, x < 0 \\ k, x = 0 \\ \frac{x}{\left| x \right|}, x > 0\end{cases}\] 

\[\Rightarrow f\left( x \right) = \begin{cases}\frac{1 - \cos2x}{2 x^2}, x < 0 \\ k, x = 0 \\ 1, x > 0\end{cases}\]

We have
(LHL at x = 0) = 

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

\[= \lim_{h \to 0} \left( \frac{1 - \cos2\left( - h \right)}{2 \left( - h \right)^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{1 - \cos2h}{2 h^2} \right)\]
\[ = \frac{1}{2} \lim_{h \to 0} \left( \frac{2 \sin^2 h}{h^2} \right)\]
\[ = \frac{2}{2} \lim_{h \to 0} \left( \frac{\sin^2 h}{h^2} \right)\]
\[ = \frac{2}{2} \lim_{h \to 0} \left( \frac{\text{ sin } h}{h} \right)^2 \]
\[ = 1 \times 1\]

(RHL at x = 0) = 

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

\[= \lim_{h \to 0} \left( 1 \right) = 1\]

Also,

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

If f(x) is continuous at x = 0, then

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

\[\Rightarrow 1 = 1 = k\]

Hence, the required value of k is 1.