Find the limit as $n \rightarrow \infty $ of
$\frac{\sin(n)-n}{n^3}$
What is a good starting point for this equation?
Help please.
Find the limit as $n \rightarrow \infty $ of
$\frac{\sin(n)-n}{n^3}$
What is a good starting point for this equation?
Help please.
$\lim_{n \to \infty}\frac{\sin n - n}{n^3}=\lim_{n \to \infty}\frac{\sin n}{n^3}-\frac{1}{n^2}$
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
$\lim_{n \to \infty}\frac{\sin n - n}{n^3}=0-0=0$
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 $|\sin x| \le 1$.
Hint: $\frac{\frac{sin(n)}{n}-1}{n^2}$ and the Squeeze theorem.