Answer
The limit does not exist
Work Step by Step
Divide both the numerator and denominator with x
$...=\displaystyle \lim_{x\rightarrow-\infty}\frac{1}{2\cdot\dfrac{\sin x}{x}}=\frac{1}{2\cdot\displaystyle\lim_{x\rightarrow-\infty}\frac{\sin x}{x}}=...$
Knowing that
$1\geq\sin x\geq-1,$
we divide with a NEGATIVE NUMBER x,
$\displaystyle \frac{1}{x}\leq\dfrac{\sin x}{x}\leq-\frac{1}{x}.$
Since $\displaystyle \lim_{x\rightarrow-\infty}\frac{1}{x}=\lim_{x\rightarrow-\infty}\frac{-1}{x}=0 ,$
by the Squeeze thorem (Theorem 1.8),
$0\displaystyle \leq\lim_{x\rightarrow-\infty}\frac{\sin x}{x}\leq 0,$
meaning that $\displaystyle \lim_{x\rightarrow-\infty}\frac{\sin x}{x}=0.$
$\displaystyle \frac{\sin x}{x}$ approaches 0 by alternating between the positive and negative
So,
$... \displaystyle \frac{1}{2\cdot\displaystyle\lim_{x\rightarrow-\infty}\frac{\sin x}{x}}$ = alternates between $+\infty$ and $-\infty$
(does not exist)
The limit does not exist.
