calculus - How to calculate $ lim_{xto 4} (frac{frac{pi}{6} - arcsin(frac{sqrt{x}}{4})}{sqrt[3]{2x-7}-1}) $ without the rule of L'Hôpital?

$$
\lim_{x\to 4} (\frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{\sqrt[3]{2x-7}-1})
$$

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.

Answer

Well, let's put $u=\arcsin(\sqrt{x} /4)$ so that as $x\to 4$ we have $u\to\pi/6$ and $x=16\sin^2u$. The expression under limit can be written as $$\frac{1}{16}\cdot\frac{\pi/6-u}{1/2-\sin u}\cdot\frac{1}{1/2+\sin u}\cdot\frac{4-x}{\sqrt[3]{2x-7}-1 }$$ The desired limit is then equal to $$\frac{1}{16}\cdot\frac{1}{\cos(\pi/6)}\cdot 1\cdot\lim_{x\to 4}\frac{4-x}{\sqrt[3]{2x-7}-1}$$ The above simplifies to $$\frac{1}{8\sqrt{3}}\lim_{t\to 1}\dfrac{1-t^3}{2(t-1)}$$ using substitution $t=\sqrt[3]{2x-7}$. Thus the desired limit is $-\sqrt{3}/16$.