$\lim_\limits{x\to 0^{+}} \frac{\arcsin(1-\{x\})\arccos(1-\{x\})}{\sqrt{2\{x\}}(1-\{x\})}$

Question

Find $$\lim_{x\to 0^{+}} \frac{\arcsin(1-\{x\})\arccos(1-\{x\})}{\sqrt{2\{x\}}(1-\{x\})}$$ where $\{\cdot\}$ stands for the fractional part function, without using the L'Hopital Rule.

Answer: $\frac{\pi}{2}$

Here's what I did:

Since $0<0^{+}<1\Rightarrow$ $$\lim_{x\to 0^{+}} \frac{\arcsin(1-\{x\})\arccos(1-\{x\})}{\sqrt{2\{x\}}(1-\{x\})}=\lim_{x\to 0^{+}} \frac{\arcsin(1-x)\arccos(1-x)}{\sqrt{2x}(1-x)}$$ This is of the form $\frac 00$. Also, $\lim_\limits{x\to 0^{+}}\frac{\arcsin(1-x)}{1-x}=\frac{\pi}{2}$. Now we are left with $\lim_\limits{x\to 0^{+}} \frac{\arccos (1-x)}{\sqrt{2x}}$.

This is the perfect moment to apply the LH rule but the question forbids it :(

Any hints on how to proceed further will be great. Thanks a lot!

Answer 1

Let $1-x=\cos\theta$ and we get the equivalent limit

$$\lim_{x\to0^+} \frac{\arccos(1-x)}{\sqrt{2x}} = \lim_{\theta\to 0} \frac{\arccos(\cos\theta)}{\sqrt{2-2\cos\theta}} = \lim_{\theta\to 0} \frac{|\theta|}{2\left|\sin\left(\frac{\theta}{2}\right)\right|} = 1$$

Answer 2

Use $$ \cos ^{ - 1} (1 - x) = \tan^{ - 1} \left( {\frac{{\sqrt {1 - (1 - x)^2 } }}{{1 - x}}} \right) = \tan^{ - 1} \left( {\sqrt x \frac{{\sqrt {2 - x} }}{{1 - x}}} \right) $$ and write $$ \frac{{\cos ^{ - 1} (1 - x)}}{{\sqrt {2x} }} = \frac{1}{{\sqrt 2 }}\frac{{\sqrt {2 - x} }}{{1 - x}}\frac{1}{{\sqrt x \dfrac{{\sqrt {2 - x} }}{{1 - x}}}}\tan^{ - 1} \left( {\sqrt x \frac{{\sqrt {2 - x} }}{{1 - x}}} \right). $$ You can finish by noting that $$ \mathop {\lim }\limits_{z \to 0} \frac{{\tan ^{ - 1} z}}{z} = \mathop {\lim }\limits_{w \to 0} \frac{w}{{\tan w}} = \mathop {\lim }\limits_{w \to 0} \frac{w}{{\sin w}}\mathop {\lim }\limits_{w \to 0} \cos w = 1. $$