Right Hand Limit of function consisting of inverse trignometric functions I'm trying to evaluate
$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - \{x\}) \times \cos^{-1}(1 - \{x\})}{\sqrt{2\{x\}}\times(1 - \{x\})}
$$
where $\{x\}$ represents the fractional part of the number.
Clearly, for this problem $\{x\} = x$. We can also split up the limit into two parts
$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - x)}{1 -x}\times \lim \limits_{x \to 0^+} \frac{\cos^{-1}(1 - x)}{\sqrt{2x}}
$$
The limit on the left is $\frac{\pi}{2}$ which we get upon substituing the $x$ for a zero. The limit on the right is where I'm struggling.
I'm looking for a way to evaluate the limit on the right and/or a different way to approach this problem altogether.
 A: You can use L'Hopital:$$\lim_{x\to0^+}\frac{\cos^{-1}(1-x)}{\sqrt{2x}}=\sqrt2\lim_{x\to0^+}\frac{\sqrt{x}}{\sqrt{1-(1-x)^2}}=1$$

You can keep $\cos^{-1}(1-x)=m$. Then $\cos m=1-x$,$$\lim_{m\to0^+}\frac{m}{\sqrt{2(1-\cos m)}}=\lim_{m\to0^+}\frac{m}{\sqrt{4\sin^2(m/2)}}=1$$
A: For the limit on right convert the $\cos^{-1}$ to $\sin^{-1}$ to get
$$\lim_{x \to 0^+}\dfrac{\sin^{-1}(\sqrt{1-(1-x)^2})}{\sqrt{2}x}$$
which can be evaluated easily
A: For $\lim_{x \to 0^+}$ we write $x=1+h, h>0$
then
$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - \{x\}) \times \cos^{-1}(1 - \{x\})}{\sqrt{2\{x\}}\times(1 - \{x\})}$$
$$ L=\lim_{h\to 0} \frac{\sin^{-1}(1-h) \cos^{-1}(1-h)}{\sqrt{2h} (1-h)}$$
$$L=\lim_{h\to 0}\frac{\sin^{-1}(1-h)}{(1-h)} \frac{\cos^{-1}(1-h)}{\sqrt{2h}}$$
$$L=\frac{\pi}{2} \lim_{h \to 0}\frac{\cos^{-1}(1-h)}{\sqrt{2h}}$$
Use L=Hosp. now
$$L=\frac{\pi}{2} \lim_{h \to 0} \frac{2/\sqrt{1-(1-h)^2}~\sqrt{h}}{\sqrt{2}}$$
Use $(1-h)^2\to 1-2h$, then
$$L=\frac{\pi}{2} \lim_{h \to 0} \frac{2\sqrt{h}}{2\sqrt{h}}=\frac{\pi}{2}.$$
