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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.

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    $\begingroup$ If you use "\arccos", "\arcsin" and "\arctan" your post will look nicer $\endgroup$
    – user
    Commented Mar 14, 2021 at 14:22
  • $\begingroup$ will do, thanks! $\endgroup$
    – Stamatis
    Commented Mar 14, 2021 at 14:23
  • $\begingroup$ I think the correct notation is $\sin x = x + o(x^2)$, $\cos x = 1-\frac{x^2}{2} + o(x^3)$ etc. $\endgroup$ Commented Mar 14, 2021 at 14:44
  • $\begingroup$ @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 $\endgroup$
    – Stamatis
    Commented Mar 14, 2021 at 14:54

1 Answer 1

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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}

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  • $\begingroup$ Shouldn't the first row be O(x^3) by default? I have been trying to figure this out but to no avail. $\endgroup$
    – Stamatis
    Commented Mar 20, 2021 at 20:22
  • $\begingroup$ By default it should be $U(x^3)$, but since $\cos x = \cos (-x)$ it can be shown that it's actually $O(x^4)$. $\endgroup$ Commented Mar 20, 2021 at 20:28
  • $\begingroup$ Where can I find more information about U(x^3), I dont seem to be able to find any articles relating to it. What is it called? $\endgroup$
    – Stamatis
    Commented Mar 20, 2021 at 21:35
  • $\begingroup$ Tthat was a typo, I meant $O(x^3)$. $\endgroup$ Commented Mar 20, 2021 at 23:13

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