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?

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
$$
L = \lim_{x \to 0} 2 \left(\frac{1}{3}\right) - 2 \left(-\frac{1}{6}\right) = 1
$$
