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This question already has an answer here:

How do I evaluate this limit: $$\lim\limits_{x\to \infty}\frac{\sin x}{x}$$

Is it $0$? If yes, how so?

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marked as duplicate by Daniel Fischer Jan 20 '16 at 20:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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we can see that for $x>0$ we have $$-\frac{1}{x}\le\frac{\sin x}{x}\le+\frac{1}{x}$$ then by squeeze theorem you can conclude that $\lim\limits_{x\to+\infty}\frac{\sin x}{x}=0$ since $\lim\limits_{x\to+\infty}-\frac{1}{x}=\lim\limits_{x\to+\infty}+\frac{1}{x}=0$

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Hint: Use the squeeze theorem with two other limits: $\lim_{x \to \infty} 1/x$ and similarly $\lim_{x \to \infty} -1/x$, since $|\sin x| \leq 1$.

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From the boundedness of sinusoidal curves we know that

$$|\sin x \ | \le 1$$ hence $$\bigg|\frac{\sin x}{x}\bigg| \leq \frac{1}{x}$$ Using our rules for absolute inequalities, we find that $$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$$

Now $$\lim_{x \to \infty} -\frac{1}{x} = 0 = \lim_{x \to \infty} \frac{1}{x} $$

If you are unsure as to why this is true, consider the following plot of $\displaystyle f(x) = \frac{1}{x}$. The same argument for $\displaystyle f(x)= - \frac{1}{x}$, since this is simply a reflection in the $x$-axis.

(Note: I plotted it only for positive $x$)

enter image description here

Thus, from the Squeeze Theorem, we thus have that $$\lim_{x \to \infty}\frac{\sin x}{x}=0$$

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To answer this you can keep in mind:

$x/\infty=0$.

$\sin(x)$ is never undefined.

Therefor, we can logically deduce that the limit $=0$

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