Computing $\;\lim\limits_{x\to 4}\left (\frac{\frac{\pi}{6} - \arcsin\left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2x-7}-1}\right) $ without L'Hôpital? $$
\lim_{x\to 4}\left(\frac{\frac{\pi}{6} - \arcsin\left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2x-7}-1}\right)
$$
Hello! I need to solve this limit. I had solved it with the rule of L'Hôpital, but i can't without it. I tried multiplication by conjugate expression and using Special Limits. Please help me, I must solve it using only Special Limits and simple transformations. I can't use derivatives.

HINT
We have that by definition of derivative by $f(x)=\arcsin(\frac{\sqrt{x}}{4})$ and $g(x)=\sqrt[3]{2x-7}$ we have
$$\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{\sqrt[3]{2x-7}-1}=\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{x-4}\frac{x-4}{\sqrt[3]{2x-7}-1}=-\frac{f'(4)}{g'(4)}$$
After editing
In order to use first order exapansion let $x=y+4$ with $y\to 0$ and we obtain
$$\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{\sqrt[3]{2x-7}-1}=\lim_{y\to 0} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{y+4}}{4})}{\sqrt[3]{2y+1}-1}$$
then we have


*

*$\sqrt{y+4}=2(1+y/4)^\frac12=2+\frac y4+o(y)$

*$\arcsin(\frac{\sqrt{y+4}}{4})=\arcsin\left(\frac12+\frac y{16}+o(y)\right)=\frac{\pi}6+\frac{\sqrt 3}{24}y+o(y)$

*$\sqrt[3]{2y+1}=(1+2y)^\frac13=1+\frac23 y+o(y)$
then we obtain
$$\frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{y+4}}{4})}{\sqrt[3]{2y+1}-1}=\frac{-\frac{\sqrt 3}{24}y+o(y)}{\frac23 y+o(y)}=\frac{-\frac{\sqrt 3}{24}+o(1)}{\frac23 +o(1)}\to -\frac{\sqrt 3}{16}$$
and you can check the result by the first hint.
