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Question
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2kx}{x^2}, \text{ if } & x \neq 0 \\ 8 , \text{ if } & x = 0\end{cases}\] at x = 0
Solution
Given:
If f(x) is continuous at x = 0, then
\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{1 - \cos2kx}{x^2} = 8\]
\[ \Rightarrow \lim_{x \to 0} \frac{2 k^2 \sin^2 kx}{k^2 x^2} = 8\]
\[ \Rightarrow 2 k^2 \lim_{x \to 0} \left( \frac{\ sinkx}{kx} \right)^2 = 8\]
\[ \Rightarrow 2 k^2 \times 1 = 8\]
\[ \Rightarrow k^2 = 4\]
\[ \Rightarrow k = \pm 2\]
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