A. $\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin (6 \theta)}=$

Step 1 ：Given the limit problem: \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin (6 \theta)}\)

Step 2 ：This limit results in an indeterminate form (0/0) as \(\theta\) approaches 0.

Step 3 ：We can apply L'Hopital's rule, which states that if the limit of a function as x approaches a certain value results in an indeterminate form (0/0 or ∞/∞), then the limit of that function is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

Step 4 ：Let's find the derivatives of the numerator and the denominator. The derivative of \(\sin \theta\) (numerator) is \(\cos \theta\) and the derivative of \(\sin (6\theta)\) (denominator) is \(6\cos (6\theta)\).

Step 5 ：Applying L'Hopital's rule, we get \(\lim _{\theta \rightarrow 0} \frac{\cos \theta}{6\cos (6 \theta)}\)

Step 6 ：As \(\theta\) approaches 0, \(\cos \theta\) approaches 1 and \(\cos (6\theta)\) also approaches 1. Therefore, the limit is \(\frac{1}{6}\).

Step 7 ：Final Answer: \(\boxed{\frac{1}{6}}\)