Solving a limit by two methods with different results I'm considering this limit
$$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$
My first attempt was using the following equivalent infinitesimals
$$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^{x^2} - 1 \sim x^2 \,\,\,\,\text{when}\,\,\,\,x\rightarrow 0,$$
and then
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2xe^{x^2}-2x+\frac{2}{3}x^3}=\lim_{x\to 0}\frac{x^4}{4(e^{x^2}-1)+\frac{4}{3}x^2}\\
&=\lim_{x\to 0}\frac{x^4}{4x^2+\frac{4}{3}x^2}=\lim_{x\to 0}\frac{3}{16}x^2=0.
\end{align*}
Later, inspecting this other approach combining infinitesimals and the Taylor expansions
\begin{align*}
e^{x^2} &= 1+x^2+\frac{x^4}{2}+o(x^4),\\
\sin 2x &= 2x - \frac{4x^3}{3} + \frac{4x^5}{15} + o(x^5),\\
e^{x^2}\sin 2x &= 2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5),
\end{align*}
I get this other result
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5)-2x+\frac{2}{3}x^3}\\
&=\lim_{x\to 0}\frac{x^5/2}{-x^5/15 + o(x^5)}=-\frac{15}{2}.
\end{align*}
Can someone help me to identify what am I getting wrong? 
Thanks.

Your first method is correct and the answer is indeed $0$. The problem with your second approach is that the Taylor series of $e^{x^2}\sin(2x)$ centered at $0$ begins with $2x\color{red}+\frac23x^3$.
