<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] Here is a solution to the EDITED problem asked by OP. I will make use of 
some standard limits only, without l'Hospital rule nor Taylor series.
First, some simple transformations are required:
\begin{eqnarray*}
\frac{1}{\sin x\arctan x}-\frac{1}{\tan x\arcsin x} &=&\frac{1}{\sin
x\arctan x}-\frac{\cos x}{\sin x\arcsin x} \\
&=&\frac{1}{\sin x}\left( \frac{1}{\arctan x}-\frac{\cos x}{\arcsin x}%
\right)  \\
&=&\frac{1}{\sin x}\left( \frac{\arcsin x-\cos x\arctan x}{\arctan x\arcsin x%
}\right) .
\end{eqnarray*}
Now, we re-write this expression using the expressions involved in standard
limits as follows:
\begin{eqnarray*}
&=&\frac{1}{\sin x}\left( \frac{\arcsin x-x+x-\arctan x+\arctan x-\cos
x\arctan x}{\arctan x\arcsin x}\right)  \\
&=&\frac{1}{\sin x}\left( \frac{\left( \arcsin x-x\right) +\left( x-\arctan
x\right) +\arctan x\left( 1-\cos x\right) }{\arctan x\arcsin x}\right)  \\
&=&\frac{x}{\sin x}\frac{x}{\arctan x}\frac{x}{\arcsin x}\left( \frac{\left(
\arcsin x-x\right) +\left( x-\arctan x\right) +\arctan x\left( 1-\cos
x\right) }{x^{3}}\right)  \\
&=&\frac{x}{\sin x}\frac{x}{\arctan x}\frac{x}{\arcsin x}\left( \frac{%
\arcsin x-x}{x^{3}}+\frac{x-\arctan x}{x^{3}}+\frac{\arctan x}{x}\left( 
\frac{1-\cos x}{x^{2}}\right) \right) 
\end{eqnarray*}
Standard limits used are
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1,\ \ \ \ \lim_{x\rightarrow 0}%
\frac{\arcsin x}{x}=1,\ \ \ \ \lim_{x\rightarrow 0}\frac{\arctan x}{x}=1 \\
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2},\ \ \ \
\lim_{x\rightarrow 0}\frac{\arcsin x-x}{x^{3}}=\frac{1}{6},\ \ \ \ \
\lim_{x\rightarrow 0}\frac{x-\arctan x}{x^{3}}=\frac{1}{3}
\end{eqnarray*}
Therefore,
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{1}{\sin x\arctan x}-\frac{1}{\tan x\arcsin
x}\right) =1\cdot 1\cdot 1\left( \frac{1}{6}+\frac{1}{3}+1\cdot \left( \frac{%
1}{2}\right) \right) =1.
\end{equation*}
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