Approximation of $\sin\left(\frac{x}{n}\right)$ I wrote in my analysis notes the following: $\sin\left(\dfrac{x}{n}\right) = -\dfrac{x}{n} + \omicron\left(\left| \dfrac{x}{n} \right|\right)$. 
I'm guessing it comes from Taylor's formula but I don't understand how they got that result.
Also, it said: $\sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + \omicron\left(\left| \dfrac{x^2}{n^2} \right|\right)$ 
Do we simply assume that since $ \omicron\left(\left| \dfrac{x}{n} \right|\right) \rightarrow 0 \implies$ $(\sin\left(\dfrac{x}{n}\right) = -\dfrac{x}{n} + \omicron\left(\left| \dfrac{x}{n} \right|\right))^2 =  \sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + \omicron\left(\left| \dfrac{x^2}{n^2} \right|\right)$

The taylor series for $\sin y$ around $y=0$ is 
$$
\sin y=y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}+\cdots.
$$
Plug in $y=x/n$ to get
$$
\sin\frac{x}{n}=\frac{x}{n}\underbrace{-\frac{x^{3}}{n^{3}3!}+\frac{x^{5}}{n^{5}5!}+\cdots.}_{o(|\frac{x}{n}|)}
$$
