<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] If you want to go through L'Hospital you may separate $\frac{1}{\cos x}$ the limit of which is $1$ for $x \to 0$:
$$\frac{\tan x - \sin x}{x^3} = \frac 1{\cos x}\cdot \frac{\sin x - \sin x \cos x}{x^3}$$
So, you only need to calculate the limit for $x \to 0$ of
$$\frac{\sin x - \sin x \cos x}{x^3}= \frac{\sin x - \frac 12\sin 2x }{x^3}$$ $$\stackrel{3\times L'Hosp.}{\sim}\frac{-\cos x+4\cos 2x}{6}\stackrel{x \to 0}{\longrightarrow} \frac 12$$
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