Explaining $\lim_{x\to0^+}\frac{\sin^{-1}x^{1/2}}{\cos(x^{1/4})-1}\approx\frac{x^{1/2}}{-\frac12(x^{1/4})^2}\to -2$

Question

Can you explain to me if and why this is correct? $$\lim_{x\to0^+}\frac{\sin^{-1}x^{1/2}}{\cos(x^{1/4})-1}\approx\frac{x^{1/2}}{-\frac12(x^{1/4})^2}\to -2$$

I don’t understand how to approx these functions and I don’t understand if I have to memorized them or there is a simple way to approx trigonometrics functions in every case.

Answer 1

Use the infinitesimal equivalences. Note that $\sin (x) \approx 0$ when $x\approx 0$ (then so does $\sin^{-1}$ as $\sin (0)=0$) and that $\cos (x) -1 \approx -x^{2}/2$ when $x\approx 0$. These equivalences come from the Taylor series of $\sin$ and $\cos$.