Limit of an oscillating function over an unbounded function

Question

If we have a function $f(x)$ that is equal to $\frac{g(x)}{h(x)}$, then the limit laws tell us that $\lim \limits_{x \to a} f(x) = \frac{\lim \limits_{x \to a} g(x)}{\lim \limits_{x \to a} h(x)}$. The limit of an oscillating function as $x$ goes to $\infty$ does not exist, but a limit of any finite number over something that increases without bound is zero. So what happens if we have a function such as $f(x) = \frac{\sin x}{x}$ where the top part oscillates between finite numbers and the bottom increases without bound? Would we say that $\lim \limits_{x \to \infty} f(x)$ does not exist, or that it equals zero?

Answer 1

For $x>0$,$$ -\frac{1}{x}\leq \frac{\sin(x)}{x} \leq \frac{1}{x}$$

$$\lim_{x \rightarrow \infty} -\frac{1}{x}\leq \lim_{x \rightarrow \infty}\frac{\sin(x)}{x} \leq \lim_{x \rightarrow \infty}\frac{1}{x}$$

Hence by squeeze theorem,

$$ \lim_{x \rightarrow \infty}\frac{\sin(x)}{x}=0$$

Use the same trick for general function.