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 (90)}\\) 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 (90)} = \square\\) (type an integer or a simplified fraction.) \\(\bigcirc\\) b. the limit does not exist.

Explanation:

Step1: Analyze the denominator

First, we know that \(\sin(90^\circ)\) (assuming the angle is in degrees, or if in radians, \(\sin(\frac{\pi}{2}) = 1\)) is a constant. The value of \(\sin(90^\circ)=1\) (in degrees) or \(\sin(\frac{\pi}{2}) = 1\) (in radians). So the denominator is a constant, not dependent on \(\theta\).

Step2: Evaluate the limit of the numerator

We know that \(\lim_{\theta
ightarrow0}\sin\theta=\sin(0) = 0\) (using the continuity of the sine function, since the sine function is continuous everywhere, so \(\lim_{x
ightarrow a}\sin x=\sin a\)).

Step3: Evaluate the overall limit

Since the denominator \(\sin(90)\) is a constant (equal to 1) and the numerator \(\sin\theta\) approaches 0 as \(\theta
ightarrow0\), we can use the limit rule for a quotient where the denominator is a non - zero constant. If \(c\) is a constant and \(\lim_{x
ightarrow a}f(x)=L\), then \(\lim_{x
ightarrow a}\frac{f(x)}{c}=\frac{\lim_{x
ightarrow a}f(x)}{c}\).

Here, \(f(\theta)=\sin\theta\), \(c = \sin(90)=1\), and \(\lim_{\theta
ightarrow0}\sin\theta = 0\). So \(\lim_{\theta
ightarrow0}\frac{\sin\theta}{\sin(90)}=\frac{\lim_{\theta
ightarrow0}\sin\theta}{\sin(90)}=\frac{0}{1}=0\).

Answer:

A. \(\lim\limits_{\theta
ightarrow0}\frac{\sin\theta}{\sin(90)}=\boxed{0}\)