Evaluate $\lim _{x \to 0} [2x^{-3}(\sin^{-1}x - \tan^{-1}x )]^{2x^{-2}}$=?

Question

$$\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]^{\frac{2}{x^2}}$$

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 $g(x) = 2/x^2$

Answer 1

By series expansion of the arcsine and arctangent, we get $$\arcsin x-\arctan x = \frac12x^3 - \frac18 x^5 + o(x^5) $$

So you're looking for $$ \begin{align}\lim_{x\to 0} [1-\tfrac14x^2+o(x^2)]^{2/x^2} &= \lim_{x\to 0} \exp\left(\frac{2}{x^2}\log(1-\tfrac14x^2+o(x^2))\right) \\&= \lim_{x\to 0} \exp\left(\frac{2}{x^2}\bigl(-\tfrac14x^2+o(x^2)\bigr)\right) = \cdots \end{align} $$

Answer 2

Recall that $$\lim_{x \to \infty} \left( 1 + \frac{k}{x}\right)^x = e^k.$$ We can add $o(x^{-1})$ terms to the parentheses without changing this. If we want to evaluate, for example, $$\lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{k_2}{x^{1 + \eta}}\right)$$ where $\eta > 0$, then note that for any $\epsilon$, we can choose some large $X$ such that $x > X$ implies $$\left|\frac{k_2}{x^\eta}\right| < \epsilon$$ and thus $$e^{k-\epsilon} = \lim_{x \to \infty} \left( 1 + \frac{k-\epsilon}{x}\right) \leq \lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{k_2}{x^{1 + \eta}}\right) \leq \lim_{x \to \infty} \left( 1 + \frac{k+\epsilon}{x}\right) = e^{k+\epsilon}.$$ But since $\epsilon$ was arbitrary, this shows $$\lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{k_2}{x^{1 + \eta}} \right) = e^k.$$

Now by Taylor expansions, for $x > 0$, $$\frac{x^3}{2} - \frac{x^5}{8} < \arcsin x - \arctan x < \frac{x^3}{2} - \frac{x^5}{8} + \frac{3x^7}{16} $$ so $$\lim_{x \to 0^+} \left(1 - \frac{x^2}{4} \right)^\frac{2}{x^2} \leq \lim_{x \to 0^+} \left[ \frac{2}{x^3} \left( \arcsin x - \arctan x\right) \right]^\frac{2}{x^2} \leq \lim_{x \to 0^+} \left(1 - \frac{x^2}{4} + \frac{3x^4}{16} \right)^\frac{2}{x^2}.$$ By substituting $y = 2/x^2$, the outer limits are made into $$\lim_{y \to \infty} \left( 1 - \frac{1}{2y} \right)^y$$ and $$\lim_{y \to \infty} \left( 1 - \frac{1}{2y} + \frac{3}{4y^2} \right)^y$$ both of which equal $\boxed{e^{-1/2}}$.

The $x \to 0^-$ limit can be treated very similarly, the only caveat being that the inequalities are reversed.