Question

If  \[f\left( x \right) = \begin{cases}\frac{1 - \cos kx}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\text{is continuous at x} = 0, \text{ find } k .\]

Answer 1

Given: 

\[f\left( x \right) = \binom{\frac{1 - \ coskx}{x\ sinx}, x \neq 0}{\frac{1}{2}, x = 0}\]

If

 \[f\left( x \right)\]  is continuous at x = 0, then

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

Consider:

\[\lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{1 - \cos kx}{x \sin x} \right) = \lim_{x \to 0} \left( \frac{2 \sin^2 \frac{kx}{2}}{x \sin x} \right)\]

\[\Rightarrow \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{2 \sin^2 \frac{kx}{2}}{x^2 \left( \frac{\sin x}{x} \right)} \right)\]

\[\Rightarrow \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{\frac{2 k^2}{4} \left( \sin \frac{kx}{2} \right)^2}{\left( \frac{kx}{2} \right)^2 \left( \frac{\sin x}{x} \right)} \right)\]

\[\Rightarrow \lim_{x \to 0} f\left( x \right) = \frac{2 k^2}{4} \lim_{x \to 0} \left( \frac{\left( sin\frac{kx}{2} \right)^2}{\left( \frac{kx}{2} \right)^2 \left( \frac{\sin x}{x} \right)} \right)\]

\[\Rightarrow \lim_{x \to 0} f\left( x \right) = \frac{2 k^2}{4}\left( \frac{\lim_{x \to 0} \frac{\left( \sin \frac{kx}{2} \right)^2}{\left( \frac{kx}{2} \right)^2}}{\lim_{x \to 0} \frac{\sin x}{x}} \right)\]

\[\Rightarrow \lim_{x \to 0} f\left( x \right) = \frac{2 k^2}{4} \times 1 = \frac{k^2}{2}\]

 From equation (1), we have

\[\frac{k^2}{2} = f\left( 0 \right)\]

\[\Rightarrow \frac{k^2}{2} = \frac{1}{2}\]
\[ \Rightarrow k = \pm 1\]