<s>[INST] Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit: 
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rightarrow 0} \frac{\sin x}{x^{3} \cos x} - \lim_{x \rightarrow 0} \frac{ \sin x}{x^3},$$
Then I got stucked, could anyone help me in solving it?   
 [/INST] If we let $A_n$ be the up/down numbers we have:
$$\tan x = \sum_{n=0}^\infty\frac{A_{2n+1}}{(2n+1)!}x^{2n+1}$$
$$\sin x = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
Thus the first few coefficients of $\tan x - \sin x$ are:
$$0 + \frac{1}{2}x^3 + \frac{1}{8}x^5 + \cdots$$
Thus if we divide by $x^3$ we get a constant coefficient of $\dfrac{1}{2}$ and everything else vanishes as $x \to 0$.
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