The Taylor expansion of arctan by integrating

1. With $u=-t^2$ in $\frac1{1-u}=1+u+u^2+u^3+o(u^3)$:
$$\frac1{1+t^2}=1-t^2+t^4-t^6+o(t^6).$$

2. Since $\arctan(0)=0$ and $\arctan'(t)=\frac1{1+t^2}$, integrating term by term,
$$\arctan x=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+o(x^7).$$

3. $\arctan x-x=-\frac{x^3}3+o(x^3)$, so the limit is $-\dfrac13$.

Remark. Evaluating at $x=1$ formally gives Leibniz's formula $\frac\pi4=1-\frac13+\frac15-\cdots$. The expansion above does not prove it (it is a local statement at $0$), but it points to it.