If $\lim_{x \to \frac{\pi}{2}} \frac{\sin x}{x}$ exists finitely, then the value of $a$ is
Step-by-Step Solution
Key Concept: For a limit of the form $\frac{\sin f(x)}{g(x)}$ to exist finitely at a removable singularity, the argument of sine must vanish at the point.
For the limit $\lim_{x \to 0} \frac{\sin(12\cos x-a)}{x^2}$ to exist finitely, the numerator must approach 0 as $x \to 0$, which requires $\sin(12\cos(0) - a) = \sin(12 - a) = 0$. This means $12 - a = 0$, so $a = 0$ or $a = n\pi$ for integer $n$. For this case, $a = 0$ and with further analysis of the limit form, we get $2\cos(0) - a = 2 - 0 = 2$.
Correct Answer: 2