Question

use the relation \\( \lim \limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \\) to determine the limit.\\( \lim \limits_{\theta \to 0} \frac{9 \sin \sqrt{6} \theta}{\sqrt{6} \theta} \\)

Explanation:

Step1: Identify the substitution

Let \( u = \sqrt{6}\theta \). As \( \theta \to 0 \), \( u \to 0 \) because \( \sqrt{6}\theta \) approaches 0 when \( \theta \) does.

Step2: Rewrite the limit

The given limit is \( \lim_{\theta \to 0} \frac{9\sin\sqrt{6}\theta}{\sqrt{6}\theta} \). We can factor out the constant 9: \( 9\lim_{\theta \to 0} \frac{\sin\sqrt{6}\theta}{\sqrt{6}\theta} \). Now, substitute \( u = \sqrt{6}\theta \), so the limit becomes \( 9\lim_{u \to 0} \frac{\sin u}{u} \).

Step3: Apply the known limit

We know that \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \) (from the given relation \( \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \), here \( u \) is just a variable, so it holds for \( u \) as well). So we substitute this value into our expression: \( 9\times1 \).

Answer:

\( 9 \)