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(A) For a function $f(\alpha)$ to be continuous at $\alpha=0$,the limit of the function as $\alpha 	o 0$ must equal the value of the function at $\al

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$1/2$

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(A) For a function $f(\alpha)$ to be continuous at $\alpha=0$,the limit of the function as $\alpha 	o 0$ must equal the value of the function at $\alpha=0$. Given $f(0) = k$. We need to evaluate $\lim_{\alpha 	o 0} f(\alpha) = \lim_{\alpha 	o 0} \frac{1-\cos 6 \alpha}{36 \alpha^2}$. Using the trigonometric identity $1-\cos 	heta = 2 \sin^2(	heta/2)$,we have $1-\cos 6 \alpha = 2 \sin^2(3 \alpha)$. Substituting this into the limit: $\lim_{\alpha 	o 0} \frac{2 \sin^2(3 \alpha)}{36 \alpha^2} = \lim_{\alpha 	o 0} \frac{2}{36} \left( \frac{\sin 3 \alpha}{\alpha} ight)^2$. Using the standard limit $\lim_{x 	o 0} \frac{\sin ax}{x} = a$,we get $\lim_{\alpha 	o 0} \frac{1}{18} (3)^2 = \frac{9}{18} = \frac{1}{2}$. Since the function is continuous,$k = 1/2$.

If the function $f(\alpha) = \begin{cases} \frac{1-\cos 6 \alpha}{36 \alpha^2}, & \alpha \neq 0 \\ k, & \alpha=0 \end{cases}$ is continuous at $\alpha=0$,then $k$ is equal to . . . . . . .

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