<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] Well you got
$\lim_{x\to 0}\frac{\sec^2x-\cos x}{3x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{\frac{1}{\cos^2x}-\cos x}{x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{1-\cos^3x}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1+\cos x+\cos^2x)}{\cos^2x}\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\frac{1}{3}\frac{(1+1+1)}{1}\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\frac{1}{3}3\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
Multiplying the numerator and denominator of the limit by $(1+\cos x)$ we get
$\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x)}{x^2}\frac{1}{(1+\cos x)}$
$\lim_{x\to 0}\frac{(1-\cos^2x)}{x^2}\lim_{x\to 0}\frac{1}{(1+\cos x)}$
We know $\sin^2x+\cos^2x=1$
From this we get $(1-\cos^2x)=\sin^2x$
$\lim_{x\to 0}\frac{\sin^2x}{x^2}\frac{1}{2}$
$\frac{1}{2}\lim_{x\to 0}\frac{\sin^2x}{x^2}$
$\frac{1}{2}[\lim_{x\to 0}\frac{\sin x}{x}]^2$
Using L'Hopital's rule to evaluate the limit we get,
$\frac{1}{2}[\lim_{x\to 0} \frac{\cos x}{1}]^2$
$\frac{1}{2}(1)^2$
$\frac{1}{2}$
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