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

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.

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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 $-\sqrt{3}/16$.

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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.

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  • $\begingroup$ Sorry, i haven't said it, but i can't use derivatives. $\endgroup$ – Влад Сивирин Nov 12 '18 at 18:36
  • $\begingroup$ @ВладСивирин Then we can elaborate that by standard limits by a change of variables. I can't elaborate it now. $\endgroup$ – user Nov 12 '18 at 18:38

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