How to prove $\displaystyle\lim_{x \to 0} \dfrac{\sin^{-1} x}{x} = 1$? How to prove this? Is there any geometrical proof? I have proved , btw, $\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x} = 1$ by Sandwich Theorem and little geometry.
you already have $$\lim_{t\to 0} \frac{\sin t}{t}  = 1.$$  we will make a change of variable $$\sin t = x, t = \sin^{-1} x $$  so that $$\lim_{x\to 0} \frac{\sin^{-1}(x)}{x} = \lim_{t\to 0}\frac{t}{\sin t} = 1.$$
I use l'Hôpital theorem. Here is a description