I want to rigorously prove that $\displaystyle L= \lim_{x \to 0} \frac{\sin^{-1}x}{x} =1$, without using l'hospital rule.
When we let $\sin y = x$, can I write $\displaystyle L= \lim_{y \to 0} \frac{\sin^{-1}(\sin y)}{\sin y}= \lim_{y \to 0} \frac{y}{\sin y}=1$ ?
Or, do I need an argument like
$\displaystyle L= \lim_{y\to k\pi} \frac{\sin^{-1}(\sin y)}{\sin y}= \lim_{y \to k\pi}\frac{y-k\pi}{\sin y}=\lim_{y \to k\pi}\frac{y-k\pi}{\sin (y-k\pi)}= \lim_{z \to 0} \frac{z}{\sin z}=1$?
What are the assumptions in each step above? For example, "if $y$ is close enough to $k\pi$, then $\sin y$ is one-to-one. Otherwise, the first equality above might not be true" etc.