<s>[INST] Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$ 
Evaluate  $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$

It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
 [/INST] The problem as corrected by the OP is now more symmetric.  We can consider cases characterized as
$$
L = \lim_{x \to 0} \frac{1}{f(x)g^{-1}(x)}-\frac{1}{f^{-1}(x)g(x)}
$$
where smooth $f(x) = x + \varepsilon_f x^m + o(x^{m+1}), g(x) = x + \varepsilon_g x^n + o(x^{n+1})$ with integers $m, n > 1$.  Then $f^{-1}(x) = x - \varepsilon_f x^m + o(x^{m+1}), g^{-1}(x) = x - \varepsilon_g x^n + o(x^{n+1})$, and
$$
\begin{align}
L & = \lim_{x \to 0} \frac{1}{(x+\varepsilon_fx^m)(x-\varepsilon_gx^n)} -
                     \frac{1}{(x-\varepsilon_fx^m)(x+\varepsilon_gx^n)} \\
  & = \lim_{x \to 0} \frac{2\varepsilon_gx^{n+1}-2\varepsilon_fx^{m+1}}
                          {x^4+o(x^6)} \\
  & = \lim_{x \to 0} 2\varepsilon_gx^{n-3}-2\varepsilon_fx^{m-3}
\end{align}
$$
In this case, we have $f(x) = \sin x, g(x) = \tan x$, so $\varepsilon_f = -1/6, \varepsilon_g = 1/3, m = n = 3$, and then
$$
L = \lim_{x \to 0} 2 \left(\frac{1}{3}\right) - 2 \left(-\frac{1}{6}\right) = 1
$$
</s>