Answer
(a) $\lim\limits_{x \to \infty}\frac{sin~x}{x} = 0$
(b) The graph of $\frac{sin~x}{x}$ crosses the asymptote $y = 0$ infinitely many times.
Work Step by Step
(a) $-1 \leq sin~x \leq 1~~~$ for all values of $x$
$-\frac{1}{x} \leq \frac{sin~x}{x} \leq \frac{1}{x}~~~$ for all values of $x \gt 0$
$\lim\limits_{x \to \infty}-\frac{1}{x} = \lim\limits_{x \to \infty}\frac{1}{x}= 0$
According to the Squeeze Theorem, $\lim\limits_{x \to \infty}\frac{sin~x}{x} = 0$
(b) The asymptote is $y=0$
The value of the function $~~sin~x~~$ moves continuously between $-1$ and $1$ over a period of $2\pi$, and the graph crosses the x-axis at the points $~~x = \pi~n~~$, where $n$ is an integer. As $x$ increases, the amplitude of the graph of $\frac{sin~x}{x}$ decreases, as the graph of $\frac{sin~x}{x}$ crosses the asymptote $y = 0$ infinitely many times at the points $~~x = \pi~n$
The graph below shows a snapshot of the function as $x$ increases. Note that the amplitude of the graph decreases as $x$ increases. Also note that the graph crosses the aymptote $~~y=0~~$ at the points $~~\pi~n~~$ where $n$ is an integer.
