<s>[INST] $ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
 [/INST] Simplify before applying the rule...
$$
\lim_{x\to 0} \frac{\tan x- \sin x}{x^3} = \lim_{x\to 0}\frac{\sin x - \sin x \cos x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0}\frac{1-\cos x}{x^2} = \lim_{x\to 0}\frac{1-\cos x}{x^2}
$$
and then apply the rule
$$
 \lim_{x\to 0}\frac{1-\cos x}{x^2}= \lim_{x\to 0}\frac{\sin x}{2x} = \frac 12.
$$
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