Question

use the relation (lim_{\theta \to 0} \frac{sin \theta}{\theta} = 1) to determine the limit.
(lim_{\theta \to 0} \frac{6 sin sqrt{4} \theta}{sqrt{4} \theta})

select the correct answer below and, if necessary, fill in the answer box to complete your choice.

a. (lim_{\theta \to 0} \frac{6 sin sqrt{4} \theta}{sqrt{4} \theta} = square) (type an integer or a simplified fraction.)

b. the limit does not exist.

Explanation:

Step1: Identify the substitution

Let \( u = \sqrt{4}\theta= 2\theta \). As \( \theta \to 0 \), \( u \to 0 \) (since \( 2\theta \to 0 \) when \( \theta \to 0 \)).
The given limit is \( \lim_{\theta \to 0} \frac{6\sin\sqrt{4}\theta}{\sqrt{4}\theta}=\lim_{\theta \to 0} 6\times\frac{\sin(2\theta)}{2\theta} \) (because \( \sqrt{4} = 2 \)).

Step2: Apply the limit formula

We know that \( \lim_{x \to 0}\frac{\sin x}{x}=1 \). Here, if we let \( x = 2\theta \), then as \( \theta \to 0 \), \( x \to 0 \). So \( \lim_{\theta \to 0}\frac{\sin(2\theta)}{2\theta}=\lim_{x \to 0}\frac{\sin x}{x} = 1 \).

Step3: Multiply by the constant factor

Now, the original limit is \( 6\times\lim_{\theta \to 0}\frac{\sin(2\theta)}{2\theta} \). Substituting the value of the limit we found in Step 2, we get \( 6\times1 = 6 \).

Answer:

A. \( \lim\limits_{\theta \to 0} \dfrac{6\sin \sqrt{4}\theta}{\sqrt{4}\theta}=\boxed{6} \)