Find this limit without using L'Hopital's rule or Taylor Series

I want to solve $$\lim_{x\to 1^{-}}{\frac{\sqrt[4]{\cos{\frac{\pi x}{2}}}}{\arccos(\frac{2}{\pi}\arcsin(x))}}$$ without using differentiation in any context. Basically without using L'Hopital's rule and Taylor's theorem.

I have managed to simplify it somehow by substituting $$x = -\frac{2}{\pi}y +1$$: $$\lim_{x\to 1^{-}}{\frac{\sqrt[4]{\cos{\frac{\pi x}{2}}}}{\arccos(\frac{2}{\pi}\arcsin(x))}} =$$ $$\lim_{y\to 0^{+}}{\frac{\sqrt[4]{\sin{y}}}{\arccos(\frac{2}{\pi}\arcsin(-\frac{2}{\pi}y +1))}}=$$ $$\lim_{y\to 0^{+}}{\frac{\sqrt[4]{\sin{y}}}{\sqrt[4]{y}}}\lim_{y\to 0^{+}}{\frac{\sqrt[4]{y}}{\arccos(\frac{2}{\pi}\arcsin(-\frac{2}{\pi}y +1))}} =$$ $$1 \lim_{y\to 0^{+}}{\frac{\sqrt[4]{y}}{\arccos(\frac{2}{\pi}\arcsin(-\frac{2}{\pi}y +1))}}$$

However I am having trouble simplifying further.

I am allowed to use the following facts: $$\sin{x} = x + o(x), \quad x \to 0$$ $$\tan{x} = x + o(x), \quad x \to 0$$ $$\arcsin{x} = x + o(x), \quad x \to 0$$ $$\arctan{x} = x + o(x), \quad x \to 0$$ $$\ln{(1+x)} = x + o(x), \quad x \to 0$$ $$\cos{x} = 1 - \frac{x^2}{2} + o(x^2), \quad x \to 0$$ $$a^x = 1+ x\ln{a} + o(x), \quad x \to 0$$ $$(1+x)^a = 1 + ax + o(x), \quad x \to 0$$ $$\cos{x}\sim 1, \quad x \to 0$$ $$if \quad \lim_{x \to a}{g(x) \ln{f(x)}} = A \quad then, \ \lim_{x \to a}{(f(x))^{g(x)}} = \begin{cases} e^A \ &if \ A \in R \\ \infty \ &if \ A = \infty \\ 0 \ &if \ A = -\infty \end{cases}$$

Any hints or help will be very much appreciated.

• If you use "\arccos", "\arcsin" and "\arctan" your post will look nicer
– user
Commented Mar 14, 2021 at 14:22
• will do, thanks! Commented Mar 14, 2021 at 14:23
• I think the correct notation is $\sin x = x + o(x^2)$, $\cos x = 1-\frac{x^2}{2} + o(x^3)$ etc. Commented Mar 14, 2021 at 14:44
• @AdamLatosiński Since I am using little o, from Taylors theorem: f(x-a) = ... + (etc)*(x-a)^k + o((x-a)^k). If I was using big O, f(x-a) = ... + (etc)*(x-a)^k + O((x-a)^{k+1}). Therefore ie, sinx = x^1 + o(x^1). However this is only my understanding, I am not certain Commented Mar 14, 2021 at 14:54

From $$\cos x = 1 - \frac{x^2}{2} + O(x^4) \qquad \text{for }x\to 0$$ we can conclude $$x = \arccos\Big(1-\frac{x^2}{2} + O(x^4)\Big) \qquad \text{for }x\to 0$$ $$\sqrt{2t} = \arccos\Big(1-t + O(t^2)\Big) \qquad \text{for }t\to 0^+$$ $$\arccos(1-t) = \sqrt{2t\big(1 + O(t)\big)} \qquad \text{for }t\to 0^+$$ $$\arccos(y) = \sqrt{2(1-y)\big(1 + O(1-y)\big)} \qquad \text{for }y\to 1^-$$ $$\arccos(y) = \sqrt{2(1-y)} + O\big((1-y)^\frac32\big) \qquad \text{for }y\to 1^-$$ Since $$\arcsin x = \frac{\pi}{2}-\arccos x$$ we have $$\arcsin(x) = \frac{\pi}{2} - \sqrt{2(1-x)} + O\big((1-x)^\frac32\big) \qquad \text{for }x\to 1^-$$ $$\frac{2}{\pi}\arcsin(x) = 1 - \frac{2}{\pi}\sqrt{2(1-x)} + O\big((1-x)^\frac32\big) \qquad \text{for }x\to 1^-$$ \begin{align} \arccos \big(\frac{2}{\pi}\arcsin(x)\big) &= \arccos\Big(1 - \frac{2}{\pi}\sqrt{2(1-x)} + O\big((1-x)^\frac32\big) \Big) = \\ &= \sqrt{\frac{4}{\pi}\sqrt{2(1-x)} + O\big((1-x)^\frac32\big)} + O\big((1-x)^\frac34\big) = \\ &= \sqrt{\frac{4\sqrt{2}}{\pi}} (1-x)^\frac14 + O\big((1-x)^\frac34\big) \qquad \text{for }x\to 1^- \end{align} Similarily, because $$\cos(x) = \sin(\frac{\pi}{2}-x)$$ we have $$\cos(x) = (\frac{\pi}{2}-x) + O\big((\frac{\pi}{2}-x)^3\big) \qquad \text{for } x\to \frac{\pi}{2}$$ $$\cos(\frac{\pi x}{2}) = \frac{\pi}{2}(1-x) + O\big((1-x)^3\big) \qquad \text{for } x\to 1$$ \begin{align} \sqrt[4]{\cos(\frac{\pi x}{2})} &= \sqrt[4]{\frac{\pi}{2}(1-x) + O\big((1-x)^3\big)}\\ &= \sqrt[4]{\frac{\pi}{2}(1-x)} \sqrt[4]{1+O\big((1-x)^2\big)} =\\ &= \sqrt[4]{\frac{\pi}{2}} (1-x)^\frac14 + O\big((1-x)^\frac94\big) \qquad \text{for } x\to 1^- \end{align} In total, we have \begin{align} \frac{\sqrt[4]{\cos(\frac{\pi x}{2})}}{\arccos \big(\frac{2}{\pi}\arcsin(x)\big)} &= \frac{\sqrt[4]{\frac{\pi}{2}} (1-x)^\frac14 + O\big((1-x)^\frac94\big)}{\sqrt{\frac{4\sqrt{2}}{\pi}} (1-x)^\frac14 + O\big((1-x)^\frac34\big)} = \\ &= \frac{\pi^\frac34}{2\sqrt{2}} + O\big((1-x)^\frac12\big)\end{align}
• By default it should be $U(x^3)$, but since $\cos x = \cos (-x)$ it can be shown that it's actually $O(x^4)$. Commented Mar 20, 2021 at 20:28
• Tthat was a typo, I meant $O(x^3)$. Commented Mar 20, 2021 at 23:13