Question

Find the value of k for which \[f\left( x \right) = \begin{cases}\frac{1 - \cos 4x}{8 x^2}, \text{ when}  & x \neq 0 \\ k ,\text{ when }  & x = 0\end{cases}\] is continuous at x = 0;

 

Answer 1

Given:

\[f\left( x \right) = \binom{\frac{1 - \cos4x}{8 x^2}, \text{ when }  x \neq 0}{k, \text{ when }  x = 0}\]If f(x) is continuous at x = 0, then

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{1 - \cos4x}{8 x^2} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{2 \sin^2 2x}{8 x^2} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2}{2} \lim_{x \to 0} \frac{\sin^2 2x}{4 x^2} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2}{2} \lim_{x \to 0} \left( \frac{\sin2x}{2x} \right)^2 = f\left( 0 \right)\]
\[ \Rightarrow 1 \times 1 = f\left( 0 \right)\]
\[ \Rightarrow k = 1 \left( \because f\left( 0 \right) = k \right)\]