<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] God knows why do you want to use L'Hopital rule only. Either you can use brute force Differentiation or you can simply put aside $\cos x$ as others have pointed out, because it is not causing any problem in the limit.
However, a creative way to do this limit-
Let $L= \displaystyle \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$
Now, let $x=3\theta$ , as $ x \rightarrow 0$ , $\theta \rightarrow 0$
$$L= \displaystyle \lim_{\theta \to 0 } \frac{\tan (3 \theta) - \sin(3\theta)}{(3\theta)^3} \\ = \displaystyle \lim_{\theta \to 0 } \dfrac{\frac{3 \tan(\theta)- \tan^3 (\theta)}{1-3\tan^2(\theta)} -(3\sin(\theta)-4\sin^3(\theta))}{27\theta^3} $$
Take the LCM and arrange the terms:
$$ L= \displaystyle \lim_{\theta \to 0} \dfrac{3\tan \theta -3\sin \theta - \tan^3 \theta +4\sin^3 \theta +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3(1-3\tan^2 \theta) }$$
Note that $(1-3\tan^2 \theta)$ is  just $1$ as $\theta \rightarrow 0$, So, we can shift the limit on it so separate it out. (I would appreciate if someone could write that in a better way, I can't)
This limit is now limited to
$$ L=  \displaystyle \lim_{\theta \to 0} \dfrac{3(\tan \theta -\sin \theta) +4\sin^3 \theta - \tan^3 \theta  +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3} $$
Do you feel the Deja Vu?
$$L =  \frac{3L}{27}  +\displaystyle \lim_{\theta \to 0}   \dfrac{4\sin^3 \theta - \tan^3 \theta  +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3}  \\ \implies 24L= 4 \displaystyle \lim_{\theta \to 0} \frac{\sin^3 \theta}{\theta^3} - \displaystyle \lim_{\theta \to 0}  \frac{\tan^3\theta}{\theta^3} + 9 \displaystyle \lim_{\theta \to 0} \frac{\sin \theta \tan^2 \theta }{\theta^3 } -12 \displaystyle \lim_{\theta \to 0}  \frac{\sin^3 \theta \tan^2 \theta}{\theta^3} $$
Hence,  $ L = \frac{ 4 -1 +9}{24} = \frac{1}{2}$
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