Find the limit $\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)$

Question

Find the limit $$\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)$$

\begin{align} \lim_{x\to 0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)&=\lim_{x\to 0}\left(\frac{\sin x-\arcsin x}{\sin x\times \arcsin x}\right)\\ &=\lim_{x\to 0}\frac{x}{\arcsin x} \cdot\frac{x}{\sin x}\cdot \frac{\sin x-\arcsin x}{x^2} \end{align}

Here's where I'm stuck. I know from L'Hopital's Rule that $\frac{x}{\sin x}\to 1$, but what about $\frac{x}{\arcsin x}$? Or is there any other way than using L'Hopital to solve this question?

Answer 1

$$\frac{1}{\arcsin{x}}-\frac{1}{\sin{x}}=\frac{1}{x+\frac{x^3}{6}+...}-\frac{1}{x-\frac{x^3}{6}+...}\rightarrow0.$$

Your way also helps because $$\frac{\sin{x}-\arcsin{x}}{x^2}\rightarrow\frac{\cos{x}-\frac{1}{\sqrt{1-x^2}}}{2x}\rightarrow\frac{-\sin{x}-\frac{x}{\sqrt{(1-x^2)^3}}}{2}\rightarrow0.$$

Answer 2

Hint:

$$\dfrac{\sin x-\arcsin x}{\sin x\cdot\arcsin x}=x\cdot\dfrac{\dfrac{\sin x-x}{x^3}-\dfrac{\arcsin x-x}{x^3}}{\dfrac{\sin x}x\cdot\dfrac{\arcsin x}x}$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion

Answer 3

$$L=\lim_{x\to 0} \left( \frac{1}{\sin^{-1} x}-\frac{1}{\sin x}\right)$$ Note that if $|x|$ is very small then $\sin^{-1} x=x+x^3/6+..., \sin x=x-x^3/6+... (1+x)^{k}=1+kx+...$ Then $$L=\lim_{x \to 0} \left( \frac{1}{x+x^3/6}-\frac{1}{x-x^3/6} \right)$$ $$\implies L=\lim_{x \to 0}\frac{1}{x}[(1+x^2/6)^{-1}-(1-x^2/6)^{-1}]=\lim_{x \to 0}\frac{1}{x}[(1-x^2/6)-(1+x^2/6)]=0$$