Key Concepts and Formulas

• Greatest Integer Function ($[x]$): The greatest integer less than or equal to $x$. For $x \to 0^+$, we are considering values of $x$ such that $0 < x < 1$. For these values, $[x] = 0$.
• Standard Limits:
  • $\mathop {\lim }\limits_{x \to 0} \frac{\sin^{-1} x}{x} = 1$
  • $\mathop {\lim }\limits_{y \to 0} \cos^{-1} y = \cos^{-1} 0 = \frac{\pi}{2}$
• Limit Properties: The limit of a product is the product of the limits (if they exist).

Step-by-Step Solution

Step 1: Analyze the expression as $x \to 0^+$ We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}$ As $x \to 0^+$, we have $0 < x < 1$. For these values of $x$, the greatest integer function $[x] = 0$. Substituting $[x] = 0$ into the expression, we get: $L = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {0^2}} \right).{{\sin }^{ - 1}}\left( {x - {0^2}} \right)} \over {x - {x^3}}}$ $L = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x.{{\sin }^{ - 1}}x} \over {x - {x^3}}}$

Step 2: Manipulate the expression to utilize standard limits We can rewrite the denominator as $x(1 - x^2)$. The expression becomes: $L = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x.{{\sin }^{ - 1}}x} \over {x(1 - {x^2})}}$ To use the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin^{-1} x}{x} = 1$, we can split the expression: $L = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{{\cos }^{ - 1}}x} \cdot \frac{{{{\sin }^{ - 1}}x}}{x} \cdot \frac{1}{1 - {x^2}} \right)$

Step 3: Apply the limit properties and standard limits We can now apply the limit to each factor: $L = \left( \mathop {\lim }\limits_{x \to {0^ + }} {{\cos }^{ - 1}}x \right) \cdot \left( \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\sin }^{ - 1}}x}}{x} \right) \cdot \left( \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{1 - {x^2}} \right)$ Let's evaluate each limit:

• $\mathop {\lim }\limits_{x \to {0^ + }} {{\cos }^{ - 1}}x$: As $x$ approaches $0$ from the positive side, $\cos^{-1} x$ approaches $\cos^{-1} 0$, which is $\frac{\pi}{2}$.
• $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\sin }^{ - 1}}x}}{x}$: This is a standard limit, and as $x \to 0$, this limit is $1$.
• $\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{1 - {x^2}}$: As $x \to 0$, $x^2 \to 0$, so $1 - x^2 \to 1$. Thus, the limit is $\frac{1}{1} = 1$.

Step 4: Calculate the final value of the limit Multiplying the results from Step 3: $L = \left( \frac{\pi}{2} \right) \cdot (1) \cdot (1)$ $L = \frac{\pi}{2}$

Common Mistakes & Tips

• Incorrectly evaluating $[x]$: For $x \to 0^+$, remember that $[x]=0$. If you mistakenly assume $[x]=x$ or some other value, the entire calculation will be wrong.
• Algebraic manipulation errors: Be careful when factoring the denominator and rearranging terms. Ensure you don't divide by zero or make incorrect cancellations.
• Recognizing standard limits: Familiarity with standard limits like $\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1$ is crucial for simplifying such problems efficiently.

Summary

The problem requires evaluating a limit involving the greatest integer function and inverse trigonometric functions. By first analyzing the behavior of $[x]$ as $x \to 0^+$ and simplifying the expression, we transformed the limit into a product of terms where standard limits could be applied. Specifically, we used the fact that for $x \to 0^+$, $[x]=0$, and the standard limit $\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1$. The limit of $\cos^{-1} x$ as $x \to 0$ is $\frac{\pi}{2}$, and the limit of the remaining term $\frac{1}{1-x^2}$ is $1$. Multiplying these values yields the final answer.

The final answer is \boxed{{\pi \over 2}}.