Key Concepts and Formulas

• Fractional Part: The fractional part of a number $x$, denoted by $\{x\}$, is defined as $\{x\} = x - \lfloor x \rfloor$. For $x \in (0, 1)$, $\{x\} = x$. For $x \in (-1, 0)$, $\{x\} = x+1$.
• Limits of Trigonometric Functions:
  • $\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1$
  • $\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1$
  • $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
• Inverse Trigonometric Function Properties:
  • $\cos^{-1}(1-y) \approx \sqrt{2y}$ for small $y > 0$.
  • $\sin^{-1}(y) \approx y$ for small $y$.
  • $\cos^{-1}(0) = \frac{\pi}{2}$.

Step-by-Step Solution

The function is given by $f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0$. We need to find the left-hand limit (L) and the right-hand limit (R) at $x=0$.

Step 1: Calculate the Right-Hand Limit (R)

To find the right-hand limit, we consider $x \to 0^+$. This means $x$ approaches 0 from the positive side. For $x > 0$ and very close to 0, we have $\lfloor x \rfloor = 0$, so $\{x\} = x$.

$\mathrm{R} = \lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$

Substitute $\{x\} = x$ for $x \to 0^+$: $\mathrm{R} = \lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right) \sin ^{-1}(1-x)}{x-x^3}$

We can factor the denominator as $x(1-x^2)$: $\mathrm{R} = \lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right) \sin ^{-1}(1-x)}{x(1-x^2)}$

Let's analyze the terms as $x \to 0^+$:

• As $x \to 0^+$, $x^2 \to 0^+$. Let $y = x^2$. Then $\cos^{-1}(1-y) \approx \sqrt{2y}$ for small $y > 0$. So, $\cos^{-1}(1-x^2) \approx \sqrt{2x^2} = x\sqrt{2}$ for small $x>0$.
• As $x \to 0^+$, $1-x \to 1^-$. So, $\sin^{-1}(1-x) \to \sin^{-1}(1) = \frac{\pi}{2}$.

Substituting these approximations and rearranging: $\mathrm{R} = \lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right)}{x} \cdot \lim _{x \rightarrow 0^{+}} \frac{\sin ^{-1}(1-x)}{1-x^2}$

Let's evaluate the first limit: $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right)}{x}$. Let $u = 1-x^2$. As $x \to 0^+$, $u \to 1^-$. Also, $x^2 = 1-u$, so $x = \sqrt{1-u}$ for $x>0$. This substitution is not straightforward. Instead, let's use the approximation $\cos^{-1}(1-y) \approx \sqrt{2y}$ for small $y$. So, $\cos^{-1}(1-x^2) \approx \sqrt{2x^2} = x\sqrt{2}$ for small $x>0$. $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right)}{x} = \lim _{x \rightarrow 0^{+}} \frac{x\sqrt{2}}{x} = \sqrt{2}$

Now evaluate the second limit: $\lim _{x \rightarrow 0^{+}} \frac{\sin ^{-1}(1-x)}{1-x^2}$. As $x \to 0^+$, $1-x \to 1^-$. Thus, $\sin^{-1}(1-x) \to \sin^{-1}(1) = \frac{\pi}{2}$. And $1-x^2 \to 1$. So, the second limit is $\frac{\pi/2}{1} = \frac{\pi}{2}$.

Combining the two limits: $\mathrm{R} = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi\sqrt{2}}{2} = \frac{\pi}{\sqrt{2}}$

Step 2: Calculate the Left-Hand Limit (L)

To find the left-hand limit, we consider $x \to 0^-$. This means $x$ approaches 0 from the negative side. For $x < 0$ and very close to 0, we have $\lfloor x \rfloor = -1$. Therefore, $\{x\} = x - \lfloor x \rfloor = x - (-1) = x+1$.

$\mathrm{L} = \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$

Substitute $\{x\} = x+1$ for $x \to 0^-$: $\mathrm{L} = \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-(x+1)^2\right) \sin ^{-1}(1-(x+1))}{(x+1)-(x+1)^3}$

Let's simplify the terms:

• $1-(x+1)^2 = 1-(x^2+2x+1) = 1-x^2-2x-1 = -x^2-2x = -x(x+2)$.
• $1-(x+1) = 1-x-1 = -x$.
• The denominator is $(x+1) - (x+1)^3 = (x+1)[1-(x+1)^2] = (x+1)[1-(x^2+2x+1)] = (x+1)[-x^2-2x] = (x+1)(-x)(x+2) = -x(x+1)(x+2)$.

So, the expression becomes: $\mathrm{L} = \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(-x^2-2x\right) \sin ^{-1}(-x)}{-x(x+1)(x+2)}$

As $x \to 0^-$, $-x \to 0^+$. Let $h = -x$, so $h \to 0^+$. The expression becomes: $\mathrm{L} = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}\left(h^2-2h\right) \sin ^{-1}(h)}{h(-h+1)(-h+2)}$

Let's analyze the terms as $h \to 0^+$:

• $\cos^{-1}(h^2-2h)$: As $h \to 0^+$, $h^2-2h \to 0^-$. Let $y = h^2-2h$. For small $y$, $\cos^{-1}(y) \approx \frac{\pi}{2} - y$. However, we need to be careful with the argument of $\cos^{-1}$. The domain of $\cos^{-1}$ is $[-1, 1]$. Since $h$ is small and positive, $h^2-2h$ is negative and close to 0. Let's use the property $\cos^{-1}(z) = \frac{\pi}{2} - \sin^{-1}(z)$. This is not helpful here. Consider the behavior of $\cos^{-1}(z)$ near $z=0$. As $z \to 0^-$, $\cos^{-1}(z) \to \frac{\pi}{2}$. So, $\cos^{-1}(h^2-2h) \to \cos^{-1}(0) = \frac{\pi}{2}$.

• $\sin^{-1}(h)$: As $h \to 0^+$, $\sin^{-1}(h) \approx h$.

• Denominator: $h(-h+1)(-h+2) \to h(1)(2) = 2h$.

Substituting these into the limit: $\mathrm{L} = \lim _{h \rightarrow 0^{+}} \frac{(\pi/2) \cdot h}{2h}$ $\mathrm{L} = \frac{\pi/2}{2} = \frac{\pi}{4}$

Step 3: Calculate the Value of the Expression

We are asked to find $\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)$. We found $\mathrm{L} = \frac{\pi}{4}$ and $\mathrm{R} = \frac{\pi}{\sqrt{2}}$.

$\mathrm{L}^2 = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16}$ $\mathrm{R}^2 = \left(\frac{\pi}{\sqrt{2}}\right)^2 = \frac{\pi^2}{2}$

Now substitute these values into the expression: $\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right) = \frac{32}{\pi^2}\left(\frac{\pi^2}{16} + \frac{\pi^2}{2}\right)$

Factor out $\pi^2$: $\frac{32}{\pi^2} \cdot \pi^2 \left(\frac{1}{16} + \frac{1}{2}\right) = 32 \left(\frac{1}{16} + \frac{8}{16}\right)$ $= 32 \left(\frac{9}{16}\right)$ $= 2 \cdot 9 = 18$

The calculation in the provided solution seems to have an error. Let's re-examine the right-hand limit calculation.

Re-evaluation of Right-Hand Limit (R)

$\mathrm{R} = \lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right) \sin ^{-1}(1-x)}{x\left(1-x^2\right)}$

Let's use the standard limit $\lim_{y \to 0} \frac{\cos^{-1}(1-y)}{\sqrt{2y}} = 1$. Here, $y = x^2$. So, $\cos^{-1}(1-x^2) \approx \sqrt{2x^2} = x\sqrt{2}$ for $x > 0$.

$\mathrm{R} = \lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right)}{x} \cdot \lim _{x \rightarrow 0^{+}} \frac{\sin ^{-1}(1-x)}{1-x^2}$

First limit: $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^2\right)}{x}$. Using the approximation $\cos^{-1}(1-x^2) \approx x\sqrt{2}$: $\lim _{x \rightarrow 0^{+}} \frac{x\sqrt{2}}{x} = \sqrt{2}$

Second limit: $\lim _{x \rightarrow 0^{+}} \frac{\sin ^{-1}(1-x)}{1-x^2}$. As $x \to 0^+$, $1-x \to 1$. So, $\sin^{-1}(1-x) \to \sin^{-1}(1) = \frac{\pi}{2}$. And $1-x^2 \to 1$. So, the second limit is $\frac{\pi/2}{1} = \frac{\pi}{2}$.

Thus, $\mathrm{R} = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$. This matches the original solution's R.

Re-evaluation of Left-Hand Limit (L)

$\mathrm{L} = \lim _{x \rightarrow 0^{-}} f(x)$ For $x \to 0^-$, $\{x\} = x+1$. $\mathrm{L} = \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-(x+1)^2\right) \sin ^{-1}(1-(x+1))}{(x+1)-(x+1)^3}$ $\mathrm{L} = \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(-x^2-2x\right) \sin ^{-1}(-x)}{(x+1)(1-(x+1)^2)}$ $\mathrm{L} = \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(-x^2-2x\right) \sin ^{-1}(-x)}{(x+1)(-x^2-2x)}$

Let $x = -h$, where $h \to 0^+$. $\mathrm{L} = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}\left(h^2-2h\right) \sin ^{-1}(h)}{(-h+1)(h^2-2h)}$

As $h \to 0^+$, $h^2-2h \to 0^-$. $\cos^{-1}(h^2-2h) \to \cos^{-1}(0) = \frac{\pi}{2}$. $\sin^{-1}(h) \approx h$. $-h+1 \to 1$. $h^2-2h \approx -2h$.

$\mathrm{L} = \lim _{h \rightarrow 0^{+}} \frac{(\pi/2) \cdot h}{1 \cdot (-2h)} = \frac{\pi/2}{-2} = -\frac{\pi}{4}$

The original solution stated $L = \frac{\pi}{4}$. Let's check the steps there. In the original solution, for L: $\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^2\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^2+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^2\right)}\end{aligned}$ This step is correct, where $h$ is taken as positive for the limit $x \to 0^-$, so $\{x\} = x+1$ becomes $\{-h\} = -h+1$. The argument of $\cos^{-1}$ is $1-(1-h)^2 = 1-(1-2h+h^2) = 2h-h^2$. The argument of $\sin^{-1}$ is $1-(-h+1) = 1+h-1 = h$. The denominator is $(-h+1) - (-h+1)^3 = (-h+1)[1-(-h+1)^2] = (-h+1)[1-(1-2h+h^2)] = (-h+1)[2h-h^2]$.

So, the limit should be: $\mathrm{L} = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}\left(2h-h^2\right) \sin ^{-1}(h)}{(-h+1)(2h-h^2)}$

As $h \to 0^+$, $2h-h^2 \to 0^+$. $\cos^{-1}(2h-h^2) \to \cos^{-1}(0) = \frac{\pi}{2}$. $\sin^{-1}(h) \approx h$. $-h+1 \to 1$. $2h-h^2 \approx 2h$.

$\mathrm{L} = \lim _{h \rightarrow 0^{+}} \frac{(\pi/2) \cdot h}{1 \cdot (2h)} = \frac{\pi/2}{2} = \frac{\pi}{4}$ This matches the original solution for L. My previous calculation for L had an error in the argument of $\cos^{-1}$.

Step 4: Final Calculation

We have $\mathrm{L} = \frac{\pi}{4}$ and $\mathrm{R} = \frac{\pi}{\sqrt{2}}$.

$\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right) = \frac{32}{\pi^2}\left(\left(\frac{\pi}{4}\right)^2+\left(\frac{\pi}{\sqrt{2}}\right)^2\right)$ $= \frac{32}{\pi^2}\left(\frac{\pi^2}{16}+\frac{\pi^2}{2}\right)$ $= \frac{32}{\pi^2} \cdot \pi^2 \left(\frac{1}{16}+\frac{1}{2}\right)$ $= 32 \left(\frac{1}{16}+\frac{8}{16}\right)$ $= 32 \left(\frac{9}{16}\right)$ $= 2 \times 9 = 18$

Let's re-examine the provided solution's final calculation. $\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) \\ =16+2 \\ =18$ The provided solution has $\mathrm{R}^2 = \frac{\pi^2}{2}$ and $\mathrm{L}^2 = \frac{\pi^2}{16}$, which matches our findings. However, the step: $\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) = 16+2$ This step should be: $\frac{32}{\pi^2}\left(\frac{8\pi^2}{16}+\frac{\pi^2}{16}\right) = \frac{32}{\pi^2}\left(\frac{9\pi^2}{16}\right) = 32 \cdot \frac{9}{16} = 2 \cdot 9 = 18$ The intermediate calculation $16+2$ is incorrect. It seems there was a miscalculation in the last step of the provided solution.

Let's check the calculation $32 \times \frac{1}{16} = 2$ and $32 \times \frac{1}{2} = 16$. So, $\frac{32}{\pi^2}(\frac{\pi^2}{16} + \frac{\pi^2}{2}) = \frac{32}{\pi^2} \frac{\pi^2}{16} + \frac{32}{\pi^2} \frac{\pi^2}{2} = 2 + 16 = 18$. The provided solution's arithmetic $16+2$ is incorrect, it should be $2+16$.

Correction on the Original Solution's Arithmetic

The original solution states: $\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right)$ This part is correct. Then it says: $=16+2$ This is where the error occurs. It should be: $= \frac{32}{\pi^2} \cdot \frac{\pi^2}{2} + \frac{32}{\pi^2} \cdot \frac{\pi^2}{16}$ $= 16 + 2 = 18$ The calculation itself is correct, but the terms seem swapped or the addition order was confusingly presented.

The final result of 18 is correct.

Common Mistakes & Tips

• Fractional Part Behavior: Be very careful about the definition of $\{x\}$ for $x \to 0^+$ and $x \to 0^-$. For $x \to 0^+$, $\{x\} = x$. For $x \to 0^-$, $\{x\} = x+1$.
• Approximations: When using approximations like $\cos^{-1}(1-y) \approx \sqrt{2y}$ or $\sin^{-1}(y) \approx y$, ensure the argument is indeed small and the approximation holds.
• Limit Properties: Break down the limit into simpler parts and use known standard limits. Ensure the conditions for these limits are met.
• Algebraic Simplification: Carefully simplify expressions involving fractional parts and algebraic terms before evaluating limits.

Summary

To evaluate the limits of $f(x)$ at $x=0$, we first considered the right-hand limit by substituting $\{x\}=x$ for $x \to 0^+$. This yielded $\mathrm{R} = \frac{\pi}{\sqrt{2}}$. Then, we considered the left-hand limit by substituting $\{x\}=x+1$ for $x \to 0^-$. This yielded $\mathrm{L} = \frac{\pi}{4}$. Finally, we calculated the required expression $\frac{32}{\pi^2}(\mathrm{~L}^2+\mathrm{R}^2)$, which resulted in 18.

The final answer is $\boxed{0}$.