calculus - How to show that $frac{sin(n)}{n}$ is $1$ as $n rightarrow 0$?

How to show that $\frac{\sin(n)}{n}$

is $1$ as $n \rightarrow 0$? just hint.

Answer

Maclaurin series expansion of $\sin(n)$ is,

$$\sin(n) = n - \frac{n^3}{3!} +\frac{n^5}{5!}+...$$

Hence,

$$\frac{\sin(n)}{n} = 1-\frac{n^2}{3!} + \frac{n^4}{5!}+...$$

$$\lim_{n\to 0}\frac{\sin(n)}{n} = 1$$