Calculate $\lim_\limits{x \to 0} \frac {\cos(xe^x)-\cos(xe^{-x})}{\arcsin^3x}$ $$\lim_{x \to 0} \frac {\cos(xe^x)-\cos(xe^{-x})}{\arcsin^3x}$$ 
Formula of difference $\cos$ have not helped me.
Assuming that L'Hopital is forbidden but you can use asimptotical simplifications like big and small o notations and Taylor series.
So $\arcsin$ can be easly replaced but what I have to do with nominator?
 A: Using MacLaurin's expansion and the $o(\cdot)$ notation:
$$
\arcsin x = x + o(x)
$$
$$
e^{\pm x}=1\pm x+\frac{x^2}{2}\pm\frac{x^3}{6}+o(x^3)
$$
$$
\cos\theta = 1-\frac{\theta^2}{2}+o(\theta^3)
$$
so
$$
\cos(x e^{\pm x})=\cos \left(x\pm x^2+o(x^3)\right)=\\
1-\frac{1}{2}\left( x^2\pm 2x^3+o(x^3) \right)+o(x^3)=1-\frac{x^2}{2}\mp x^3+o(x^3)
$$
and hence
$$
\cos (x e^{x} ) - \cos (x e^{-x} ) = -2x^3 + o(x^3).
$$
Putting everything back together:
$$
\lim_{x\to 0 } \frac{\cos (x e^{x} ) - \cos (x e^{-x} )}{\arcsin^3 x}=
\lim_{x\to 0 } \frac{-2x^3 + o(x^3)}{x^3 + o(x^3)}=-2.
$$
A: We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\cos(xe^{x}) - \cos(xe^{-x})}{\arcsin^{3}x}\notag\\
&= \lim_{x \to 0}\frac{2\sin((xe^{x} + xe^{-x})/2)\sin((xe^{-x} - xe^{x})/2)}{\arcsin^{3}x}\notag\\
&= \lim_{x \to 0}\frac{2\sin((xe^{x} + xe^{-x})/2)\sin((xe^{-x} - xe^{x})/2)}{x^{3}}\cdot\frac{x^{3}}{\arcsin^{3}x}\notag\\
&= -2\lim_{x \to 0}\frac{\sin(x\cosh x)\sin(x\sinh x)}{x^{3}}\notag\\
&= -2\lim_{x \to 0}\frac{\sin(x\cosh x)}{x\cosh x}\cdot\frac{x\cosh x}{x}\cdot\frac{\sin(x\sinh x)}{x\sinh x}\cdot\frac{x\sinh x}{x^{2}}\notag\\
&= -2\cdot 1\cdot 1\cdot 1\cdot 1 = -2\notag
\end{align}
We have used the following standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\sinh x}{x} = \lim_{x \to 0}\cosh x = 1$$
A: Hint: Use $\cos (xe^x) = 1 -x^2e^{2x}/2 + O(x^4), \cos (xe^{-x}) = 1 -x^2e^{-2x}/2 + O(x^4).$
A: Since, from the comments, it seems that the problem is to do an expansion of $\cos(xe^x)$, I do this as a hint, and if you need further help, tell me.
From a look at the denominator, we hope it is sufficient to expand to order three.
Expanding $xe^x$, we have
$$
xe^x=x(1+x+x^2/2+O(x^3))=x+x^2+x^3/2+O(x^4)
$$
Since $xe^x\to 0$ as $x\to 0$, we can expand $\cos$ around $0$,
$$
\cos t=1-t^2/2+O(t^4).
$$
Now insert the other expansion,
$$
\begin{split}
\cos(xe^x)&=1-\bigl(x+x^2+x^3/2+O(x^4)\bigr)^2/2+O\bigl(x+x^2+x^3/2+O(x^4))^4\bigr)\\
&=1-x^2/2-x^3+O(x^4).
\end{split}
$$
Here, in the last step, we have included all terms of order greater than $x^3$ in the ordo term.
I think you can now proceed with the other terms?
