Different answers after different methods in solving a limit 
Evaluate $$L=\lim_{x \to 0} \frac{e^{\sin(x)}-(1+\sin(x))}{(\arctan(\sin(x)))^2}$$

Method $1$: $$\frac{h^2\left(\frac{e^{h}-1}{h^2}-\frac1{h}\right)}{(\arctan(h))^2}=1^2\left(\frac{1*1}{h}-\frac1{h}\right)=\frac1{h}-\frac1{h}=0$$ Therefore $L=0$.
The identities I have used here to simplify the expression are $$\lim_{x \to 0} \frac{\arctan\left(x\right)}{x}=1$$ $$\lim_{x \to 0} \frac{a^{x}-1}{a}=\ln\left(a\right) \implies \lim_{x \to 0} \frac{e^{x}-1}{x}=\ln\left(e\right)=1$$
Method $2$: $$L=\frac{e^{h}-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(1+h+\frac{h^{2}}{2!}\right)-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(\frac{h^{2}}{2!}\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{1}{2}$$ Therefore $L=\frac12$.
I am not at all familiar with $O(n)$ notation BTW.

My (personal) favored approach is composition of series one piece at the time.
$$y=\frac{e^{\sin(x)}-(1+\sin(x))}{\big[\arctan(\sin(x))\big]^2}$$ Since the denominator is $\sim x^2$, let us use expansions to $O(x^4)$. Working one piece at the time
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$
$$e^{\sin(x)}=1+x+\frac{x^2}{2}+O\left(x^4\right)$$
$$\arctan(\sin(x))=x-\frac{x^3}{2}+O\left(x^4\right)$$
$$\big[\arctan(\sin(x))\big]^2=x^2-x^4+O\left(x^5\right)$$ Putting all together
$$y=\frac { \frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)} {x^2-x^4+O\left(x^5\right) }=\frac{1}{2}+\frac{x}{6}+O\left(x^2\right)$$ which shows not only the limit but also how it is approached.
