Answer
The value of the limit as $x$ approaches $\frac{3\pi }{2}$, is $\underset{x\to \frac{3\pi }{2}}{\mathop{\lim }}\,\sin x=\underline{-1}$
Work Step by Step
Choose some values of $x$ and compute the respective values of $f\left( x \right)=\sin x$.
Substitute $x=0$ in $f\left( x \right)$.
$f\left( 0 \right)=\sin 0=0$
Substitute $x=\frac{\pi }{2}$ in $f\left( x \right)$.
$f\left( \frac{\pi }{2} \right)=\sin \frac{\pi }{2}=1$.
Substitute $x=-\frac{\pi }{2}$ in $f\left( x \right)$.
$f\left( -\frac{\pi }{2} \right)=\sin \left( -\frac{\pi }{2} \right)=-1$.
To find the limit of the curve, first observe the graph near $x=\frac{3\pi }{2}$ from both sides.
As $x$ approaches $\frac{3\pi }{2}$ from the left, the values of $f\left( x \right)$ get closer to $-1$.
Therefore, the left-hand limit, $\underset{x\to {{\frac{3\pi }{2}}^{-}}}{\mathop{\lim }}\,\sin x=-1$.
As $x$ approaches $\frac{3\pi }{2}$ from the right, the values of $f\left( x \right)$ get closer to $-1$.
Therefore, the right-hand limit, $\underset{x\to {{\frac{3\pi }{2}}^{+}}}{\mathop{\lim }}\,\sin x=-1$.
Because, $\underset{x\to {{\frac{3\pi }{2}}^{-}}}{\mathop{\lim }}\,\sin x=-1\text{ and }\underset{x\to {{\frac{3\pi }{2}}^{+}}}{\mathop{\lim }}\,\sin x=-1$, the limit of $f\left( x \right)$ exists when $x$ approaches $\frac{3\pi }{2}$.
Therefore, $\underset{x\to \frac{3\pi }{2}}{\mathop{\lim }}\,\sin x=-1$
