limit using taylor series I keep getting an error in the expansion
$$\lim_{x\to 0 }\frac{2\exp(\sin(x))-2-x-x^2-\arctan (x) }{x^3}$$
The numerator works out as
$$\approx 2 (1+\sin(x) + 1/2 \sin^2(x)) -2-x-x^2 - (x-x^3/3+x^5/5)$$
$$\approx 2(1+x-x^3/3!+1/2(x-x^3/3!)^2 -2-2x+x^3/3-x^2 -x^5/5 $$
$$ = o(x^4)$$
so that the limit is zero. But it is supposed to be $1/3$. Where is my mistake?

Starting from $$e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+\frac{y^4}{24}+O\left(y^5\right)$$ Replace $y$ by the expansion of $\sin(x)$ to get $$e^{\sin(x)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}+O\left(x^6\right)$$ Just as quid answered while I was typing, one term was missing in the expansion.
