# What is $\lim_{x \to 0} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$, where $\{\cdot \}$ represents fractional part function?

What is $$\lim_{x \to 0} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$$, where $\{\cdot \}$ represents fractional part function?

I have made some headway with both left and right limits, but I haven't completed any yet.

For right hand limit, $\{x\} =x$. This gives $$\lim_{x \to {0^+}} \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x-x^3}$$ I don't know how to proceed after this. I tried using L'Hopital rule, but it gets way too complicated.

Also, for left hand limit, $\{x\} = x+1$. This gives $$\lim_{x \to {0^-}} \frac{\cos^{-1}(-x^2-2x) \sin^{-1}(-x)}{(1+x)(-x^2-2)}$$

I don't know how to solve this limit further. Any help is appreciated.

• try to write 1 or 2 terms of arcsin and arcos using taylor expansion about 0. That should make it pretty easy. I think 1 term is suffecient. May 3, 2016 at 16:36
• Taylor series in not in my course syllabus, so I don't know much about that. I do know the taylor series for arcsin, but if i recall correctly, it is only valid in the neighborhood of $0$, not in $1$. Also, I don't know what the taylor expansion for arcos is May 3, 2016 at 16:46
• Taylor expansion is valid about any point as long as you stay within the domain of definition of function. Here, you are approaching 1 from the left side, so writing taylor expansion about that point wouldn't be a problem. Also, google on how to write taylor expansion for any function f(x) about a point. It will be very useful in evaluating limits (if you have IIT-JEE in your mind) May 3, 2016 at 16:52
• The fractional part takes the sign of the number. Thus, $\;\{0.13\}=0.13\;,\;\;\{-0.13\}=-0.13\;$ May 3, 2016 at 17:19
• @Joanpemo No, the range of the fractional part function is $[0,1)$. It is never negative. It is always positive, and periodic with a period of $1$. Maybe you are mistaking it with something else. May 3, 2016 at 17:26

Taylor is not necessary.

$$\lim_{x\to +0} \frac{cos^{-1}(1-x^2) sin^{-1} (1-x)}{x-x^3} = \lim_{x\to +0} \frac{cos^{-1}(1-x^2)}{x-x^3} \lim_{x\to +0}sin^{-1} (1-x) = \frac{\pi}{2} \lim_{x\to +0} \frac{cos^{-1}(1-x^2)}{x-x^3} = (L'Hopital) \frac{\pi}{2} \lim_{x\to +0} \frac{2x}{\sqrt{2x^2 - x^4}*(1-3x^2)} = \frac{\pi}{2} \lim_{x\to +0} \frac{2}{\sqrt{2 - x^2}*(1-3x^2)} = \frac{\pi}{\sqrt{2}}$$

When $\;x\to0\;$ , say $\;x\in(-0.1,\,0.1)\;$ , we already have $\;x=\{x\}\;$ (see my comment below the question), so your function in fact is

$$\frac{\arccos(1-x^2)\arcsin(1-x)}{x-x^3}=\frac{\arccos(1-x^2)}x\frac{\arcsin(1-x)}{1-x}\frac1{1+x}$$

Now, with l'Hospital:

$$\lim_{x\to0}=\frac{\arccos(1-x^2)}x\stackrel{\text{l'H}}=\lim_{x\to0}\frac{2x}{\sqrt{1-(1-x^2)^2}}=\lim_{x\to0}\frac2{\sqrt{2-x^2}}=\sqrt2$$

and thus the limit is

$$\frac{\arccos(1-x^2)}x\frac{\arcsin(1-x)}{1-x}\frac1{1+x}\xrightarrow[x\to0]{}\sqrt2\cdot\frac\pi2\cdot1=\frac\pi{\sqrt2}$$