# find the limit as $n \rightarrow \infty$ of $\frac{\sin(n)-n}{n^3}$

Find the limit as $n \rightarrow \infty$ of
$$\frac{\sin(n)-n}{n^3}$$

What is a good starting point for this equation?

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This is actually an easy question; you are overthinking it. –  hardmath Apr 29 '12 at 17:52
Try breaking it up as $\frac{\sin (n)}{n^3}-\frac{1}{n^2}$. –  Alex Becker Apr 29 '12 at 17:53

$$\lim_{n \to \infty}\frac{\sin n - n}{n^3}=\lim_{n \to \infty}\frac{\sin n}{n^3}-\frac{1}{n^2}$$

Clearly, $\sin n$ may be anywhere between -1 and 1 (i.e. a finite value). However, the denominator of $n^3$ tends to infinity when $n \to \infty$. A finite value divided by something infinitely large tends to zero. Thus

$$\lim_{n \to \infty}\frac{\sin n}{n^3}=0$$

This same principle can be used to find that

$$\lim_{n \to \infty}\frac{1}{n^2}=0$$

Thus

$$\lim_{n \to \infty}\frac{\sin n - n}{n^3}=0-0=0$$

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Theorem: Let $(a_n)$ be a bounded sequence and $(b_n)$ such that $\lim_{n \to \infty} b_n = 0$, then $\lim_{n \to \infty} a_nb_n = 0$. Recall that $|\sin x| \le 1$.

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Hint: $\frac{\frac{sin(n)}{n}-1}{n^2}$ and the Squeeze theorem.

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