What's wrong with this use of Taylor's expansions? I'm trying to find the value of the following limit:
$$
\lim_{x \to 0} \frac{x^2\cos x - \sin(x\sin x)}{x^4}
$$
Which I know equals to $-\dfrac13$.
I tried to do the following:
$$
\lim_{x \to 0} \frac{x^2(1 - \frac{x^2}{2} + o(x^4)) - \sin(x(x + o(x^3)))}{x^4}\\
= \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + o(x^4) - \sin(x^2 + o(x^4)))}{x^4}\\
= \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + o(x^4) - x^2 + o(x^4)}{x^4}
= -\frac12
$$
The result is clearly wrong. I suspect the mistake to be in the expansion of $\sin (x \sin x)$ but I don't get it.
What's wrong?
 A: Hint
$$\sin(x\sin(x))=x^2-\frac{x^4}{6}+O\left(x^5\right)$$ would help.
In fact, and this is a general rule, if the denominator is $x^n$, you must develop the numerator at least to order $n$.
A: Since you have $x^4$ in the denominator, you need to expand the numerator up to $x^4$. You did that for $x^2\cos x$ but not for $\sin (x \sin x)$.
So, you need to include terms up to $x^4$ in the expansion of $\sin (x \sin x)$:
$$\sin (x \sin x) = x^2-x^4/6+o(x^6)$$
This comes for
$$\sin (x) = x-x^3/6+o(x^5)$$
$$x\sin (x) = x^2-x^4/6+o(x^6)$$
which you then feed to
$$\sin (y) = y+o(y^3)$$
A: 
I suspect the mistake to be in the expansion of $~\sin(x\sin x),$ but I don't get it. What's wrong ?

Your error consists in using the double-approximation $\sin(x\sin x)\simeq x\sin x\simeq x^2,$ by applying 
$\sin t\simeq t$ twice instead of just once, yielding the more accurate $\sin(x\sin x)\simeq x\sin x.$ The latter 
leads to $~\lim\limits_{x\to0}~\dfrac{\cos x-\dfrac{\sin x}x}{x^2}=-\dfrac13,~$ which is different from $~\lim\limits_{x\to0}~\dfrac{\cos x-\color{red}1}{x^2}=-\dfrac12.$
