<s>[INST] $\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$ $$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$$
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?
 [/INST] $\lim_{x\to 0}\frac{e^x-1}{x}=1\implies e^x\approx 1+x $when $x\to 0$,thus, $e^{\sin^2x}\approx 1+\sin^2x$ as $x\to 0$
Hence, $$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}=\lim_{x \to 0} \frac{1+ \sin^2 x - \cos x}{x^3}$$ 
Now you can easily apply L'Hopital's rule to conclude that limit doesn't exist.
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