$$\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]^{\frac{2}{x^2}}$$  How to find this limit?  My Try: I tried to evaluate this $$\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]$$ to understand the nature of the problem. I used L'Hopital. But it became too tedious to calculate.  Can anyone please give me suggestion to solve it?  Edit: I used the hint given By lab bhattacharjee. I expand the inverse functions and I got $$\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right] =1$$.  Now I think it remains to find the value of $\lim _{x \to 0}e^{f(x).g(x)}$ where $f(x) = \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]$ and $g(x) = 2/x^2$
Evaluate:  $$\lim_{x\to{0}}\bigg(\frac{2}{x^3}.(\arcsin{x}-\arctan{x})\bigg)^{2/x^2}$$  I can just expand $\arcsin{x}$ and $\arctan{x}$ using their taylor expansions, but is there any other method?
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