TITLE: Calculate $\lim_\limits{x \to 0} \frac {\cos(xe^x)-\cos(xe^{-x})}{\arcsin^3x}$
QUESTION [2 upvotes]: $$\lim_{x \to 0} \frac {\cos(xe^x)-\cos(xe^{-x})}{\arcsin^3x}$$ 
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?

REPLY [1 votes]: 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 $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\sinh x}{x} = \lim_{x \to 0}\cosh x = 1$$
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