Question

use the relation \\(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\\) to determine the limit. \\(\lim_{\theta \to 0} \frac{\sin \theta}{\sin (11\theta)}\\) select the correct answer below and, if necessary, fill in the answer box to complete your choice. \\(\bigcirc\\) a. \\(\lim_{\theta \to 0} \frac{\sin \theta}{\sin (11\theta)} = \square\\) (type an integer or a simplified fraction.) \\(\bigcirc\\) b. the limit does not exist.

Explanation:

Step1: Rewrite the limit expression

We have the limit \(\lim_{\theta \to 0}\frac{\sin\theta}{\sin(11\theta)}\). Let's rewrite it by multiplying and dividing by \(\theta\) and \(11\theta\) appropriately. We can rewrite the expression as:
\[
\lim_{\theta \to 0}\frac{\sin\theta}{\theta}\times\frac{11\theta}{\sin(11\theta)}\times\frac{1}{11}
\]

Step2: Apply the limit formula \(\lim_{x\to 0}\frac{\sin x}{x} = 1\)

We know that \(\lim_{\theta \to 0}\frac{\sin\theta}{\theta}=1\) and if we let \(x = 11\theta\), as \(\theta\to 0\), \(x\to 0\). So \(\lim_{\theta \to 0}\frac{11\theta}{\sin(11\theta)}=\lim_{x\to 0}\frac{x}{\sin x}=1\) (since \(\lim_{x\to 0}\frac{\sin x}{x} = 1\) implies \(\lim_{x\to 0}\frac{x}{\sin x}=1\)).
Now, substitute these limits into our expression:
\[
\lim_{\theta \to 0}\frac{\sin\theta}{\theta}\times\lim_{\theta \to 0}\frac{11\theta}{\sin(11\theta)}\times\frac{1}{11}=1\times1\times\frac{1}{11}
\]

Answer:

\(\frac{1}{11}\)