$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$
Attempt
This is indeterminate of the form $\frac{0}{0}$. Applying L'Hopital's rule twice results in,
$\lim_{x \to 0} \frac{e^{\sin^2 x}(\sin^2(2x) + \sin^2(4x)) + \cos x}{6x}.$
Wolfram Alpha says the two sided limit does not exist.
Question
Can I deduce the result by splitting it up into,
$\frac{1}{6}\lim_{x \to 0} \left(e^{\sin^2 x}(\sin^2(2x) + \sin^2(4x)) + \cos x \right)\left(\frac{1}{x}\right).$
And saying that the limit does not exist because $\frac{1}{x}$ does not have a two sided limit?