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Step-by-Step Solution
Step 1: Understand the expression inside the limit
We need to evaluate the limit:
$$
\lim_{x \to 0^+} \frac{\cos^{-1}\bigl(x - [x]^2\bigr)\,\sin^{-1}\bigl(x - [x]^2\bigr)}{x - x^3}.
$$
Here, $[x]$ denotes the greatest integer less than or equal to $x$. As $x \to 0^+$ (approaching 0 from the right), we have $[x] = 0$. Therefore, $[x]^2 = 0$, and the expression inside the inverse trigonometric functions becomes simply $x$:
$$
x - [x]^2 = x - 0^2 = x.
$$
Hence, for very small positive $x$, the expression simplifies to
$$
\frac{\cos^{-1}(x)\,\sin^{-1}(x)}{x - x^3}.
$$
Step 2: Rewrite the denominator
Observe that
$$
x - x^3 = x\bigl(1 - x^2\bigr).
$$
So the limit becomes
$$
\lim_{x \to 0^+} \frac{\cos^{-1}(x)\,\sin^{-1}(x)}{x\,(1 - x^2)}.
$$
Step 3: Expand or analyze the numerator for small x
For small values of $x$, we can use the fact that:
$$
\sin^{-1}(x) \approx x,
\quad
\cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x) \approx \frac{\pi}{2} - x.
$$
A more direct way is to notice that for very small $x$:
\[
\sin^{-1}(x) \approx x,
\quad
(\sin^{-1}(x))^2 \approx x^2,
\]
and
\[
\cos^{-1}(x) \,\sin^{-1}(x)
= \bigl(\tfrac{\pi}{2} - \sin^{-1}(x)\bigr)\,\sin^{-1}(x)
= \tfrac{\pi}{2}\,\sin^{-1}(x) - \bigl(\sin^{-1}(x)\bigr)^2.
\]
Hence, as $x\to 0$,
\[
\cos^{-1}(x)\,\sin^{-1}(x)
\approx \tfrac{\pi}{2}\,x - x^2.
\]
Step 4: Substitute the approximations into the limit
Using these approximations,
\[
\cos^{-1}(x)\,\sin^{-1}(x) \approx \frac{\pi}{2}x - x^2,
\]
and
\[
x - x^3 = x(1 - x^2).
\]
Thus,
\[
\frac{\cos^{-1}(x)\,\sin^{-1}(x)}{x - x^3}
\approx
\frac{\frac{\pi}{2}x - x^2}{x(1 - x^2)}
=
\frac{x\left(\frac{\pi}{2} - x\right)}{x(1 - x^2)}
=
\frac{\frac{\pi}{2} - x}{1 - x^2}.
\]
Taking the limit as $x \to 0^+$:
\[
\lim_{x\to 0^+} \frac{\frac{\pi}{2} - x}{1 - x^2}
=
\frac{\frac{\pi}{2} - 0}{1 - 0}
=
\frac{\pi}{2}.
\]
Step 5: Conclude the value of the limit
Therefore, the value of the given limit is
\[
\boxed{\frac{\pi}{2}}.
\]
This matches the provided correct answer.
