Let's find the limit of the function [tex]\(\sin \left( \frac{\pi x}{6} \right)\)[/tex] as [tex]\(x\)[/tex] approaches -1.
To determine this limit, we can directly substitute [tex]\(x = -1\)[/tex] into the function and evaluate. Since the sine function is continuous, we can use substitution for the limit:
[tex]\[
\lim_{x \to -1} \sin \left( \frac{\pi x}{6} \right) = \sin \left( \frac{\pi (-1)}{6} \right)
\][/tex]
Now, we compute the argument of the sine function:
[tex]\[
\frac{\pi (-1)}{6} = -\frac{\pi}{6}
\][/tex]
Next, we need to find the sine of [tex]\(-\frac{\pi}{6}\)[/tex]. Recall that the sine function has the property:
[tex]\[
\sin(-\theta) = -\sin(\theta)
\][/tex]
So,
[tex]\[
\sin\left(-\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right)
\][/tex]
We know from trigonometry that:
[tex]\[
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
\][/tex]
Thus,
[tex]\[
\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \to -1} \sin \left( \frac{\pi x}{6} \right) = -\frac{1}{2}
\][/tex]
Given the options:
[tex]\[
-\frac{\sqrt{3}}{2}, \quad -\frac{1}{2}, \quad \frac{1}{2}, \quad \frac{\sqrt{3}}{2}
\][/tex]
The correct answer is [tex]\(\boxed{-\frac{1}{2}}\)[/tex].
