Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of ${\lim }_{x\to 0^{+}}\frac{{\cos }^{-1}(x-[x{]}^2)\cdot {\sin }^{-1}(x-[x{]}^2)}{x-x^3}$, where $[x]$ denotes the greatest integer $\le x$ is :

Imagine you are standing on the number line, looking at the point zero. The limit $x\to 0^{+}$ is like walking toward zero from the positive side, taking tiny, microscopic steps. This little plus sign is a vital piece of information that defines our entire journey.

When we encounter the greatest integer function $[x]$ in a limit, it often acts as a gatekeeper. For any $x$ in the interval $(0,1)$, the greatest integer less than or equal to $x$ is always $0$.

This is our key. By recognizing that $x$ is trapped in this tiny interval, we can replace $[x]$ with $0$ and watch the complexity of the problem dissolve.

With $[x]=0$, our expression $\frac{{\cos }^{-1}(x-[x{]}^2)\cdot {\sin }^{-1}(x-[x{]}^2)}{x-x^3}$ transforms into something much more manageable. The term $x-[x{]}^2$ becomes $x-0^2$, which is simply $x$.

Suddenly, the inverse trigonometric functions are just ${\cos }^{-1}(x)$ and ${\sin }^{-1}(x)$. Our numerator is now ${\cos }^{-1}(x)\cdot {\sin }^{-1}(x)$.

The denominator, $x-x^3$, is a classic algebraic structure. By factoring out an $x$, we get $x(1-x^2)$. Now, our limit looks like this:

In the world of JEE Advanced, we look for patterns. We have a ${\sin }^{-1}(x)$ in the numerator and an $x$ in the denominator, which is a perfect match for the standard limit ${\lim }_{x\to 0}\frac{{\sin }^{-1}(x)}{x}=1$.

Let us isolate this part:

The first part is a standard limit that evaluates to $1$. The second part, $\frac{{\cos }^{-1}(x)}{1-x^2}$, is no longer indeterminate. As $x$ approaches $0$, the numerator ${\cos }^{-1}(0)$ becomes $\frac{\pi }{2}$, and the denominator $1-0^2$ becomes $1$.

We are left with a simple multiplication:

It is a beautiful result. What started as a daunting expression involving step functions and inverse trigonometry collapsed into a clean, elegant constant.

This problem teaches us that even the most intimidating mathematical expressions often have a simple, logical core. Always look for the neighborhood of the limit, simplify the functions, and trust in the standard limits you have mastered. The final answer is $\frac{\pi }{2}$.