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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?

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  • $\begingroup$ Siong's answer explains it properly, but the rule-of-thumb is that "bounded over unbounded goes to zero". I.e. it's the fact that $\sin$ stays "between finite numbers", not the fact that it oscillates, that makes the overall limit zero. (E.g. $g(x) = x^2\sin(x)$ also oscillates, but your argument wouldn't work there.) $\endgroup$ Apr 2 '17 at 15:55
  • $\begingroup$ @user275313 Right, that's why I said "oscillates between finite numbers." My question only really makes sense when both are true. $\endgroup$
    – RothX
    Apr 2 '17 at 22:27
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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.

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