Key Concepts and Formulas

• Maclaurin Series Expansions: We will use the Maclaurin series expansions for $\sin^{-1}x$ and $\tan^{-1}x$ around $x=0$.
  • $\sin^{-1}x = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \dots$
  • $\tan^{-1}x = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \dots$
• Limit Evaluation: When evaluating limits of the form $\frac{0}{0}$, L'Hôpital's Rule or series expansions can be used. In this case, series expansion is more direct.
• Algebraic Manipulation: Basic algebraic operations will be used to simplify the expression after substituting the series expansions.

Step-by-Step Solution

Step 1: Identify the indeterminate form. As $x \to 0$, both the numerator ($\sin^{-1}x - \tan^{-1}x$) and the denominator ($3x^3$) approach 0. Thus, the limit is of the indeterminate form $\frac{0}{0}$. This suggests using series expansions or L'Hôpital's Rule.

Step 2: Substitute the Maclaurin series expansions for $\sin^{-1}x$ and $\tan^{-1}x$. We use the first few terms of the Maclaurin series for $\sin^{-1}x$ and $\tan^{-1}x$ as they are sufficient for evaluating the limit involving $x^3$. $\sin^{-1}x = x + \frac{x^3}{6} + O(x^5)$ $\tan^{-1}x = x - \frac{x^3}{3} + O(x^5)$ Substituting these into the limit expression: $L = \mathop {\lim }\limits_{x \to 0} {{{(x + \frac{x^3}{6} + O(x^5))) - (x - \frac{x^3}{3} + O(x^5))}} \over {3{x^3}}}$

Step 3: Simplify the numerator. Combine the terms in the numerator. Notice that the $x$ terms cancel out, and we are left with terms involving $x^3$. L = \mathop {\lim }\limits_{x \to 0} {{{x + \frac{x^3}{6} - x + \frac{x^3}{3} + O(x^5)}}} \over {3{x^3}}} L = \mathop {\lim }\limits_{x \to 0} {{(\frac{x^3}{6} + \frac{x^3}{3}) + O(x^5)}} \over {3{x^3}}} L = \mathop {\lim }\limits_{x \to 0} {{(\frac{1}{6} + \frac{1}{3})x^3 + O(x^5)}} \over {3{x^3}}}

Step 4: Combine the coefficients of $x^3$ in the numerator. $\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$ So the numerator becomes $\frac{1}{2}x^3 + O(x^5)$.

Step 5: Substitute the combined numerator back into the limit expression and evaluate. L = \mathop {\lim }\limits_{x \to 0} {{(\frac{1}{2})x^3 + O(x^5)}} \over {3{x^3}}} Divide each term in the numerator by $3x^3$: L = \mathop {\lim }\limits_{x \to 0} {{(\frac{1}{2})x^3 \over 3x^3} + \frac{O(x^5)}{3x^3}}} L = \mathop {\lim }\limits_{x \to 0} {{1 \over 6} + O(x^2)}} As $x \to 0$, the $O(x^2)$ terms go to zero. $L = {1 \over 6}$

Step 6: Calculate the value of (6L + 1). Now that we have found the value of $L$, we can substitute it into the expression $6L + 1$. $6L + 1 = 6 \times \left(\frac{1}{6}\right) + 1$ $6L + 1 = 1 + 1$ $6L + 1 = 2$

Common Mistakes & Tips

• Incorrect Series Expansion: Ensure you use the correct Maclaurin series for $\sin^{-1}x$ and $\tan^{-1}x$. A common mistake is mixing up coefficients or signs.
• Ignoring Higher-Order Terms: When dealing with limits involving $x^n$, it's crucial to expand the functions up to at least the $x^n$ term. In this case, we needed the $x^3$ terms to find a non-zero limit. The $O(x^5)$ terms, and higher, become negligible as $x \to 0$ after division by $x^3$.
• Algebraic Errors: Be careful with arithmetic when combining fractions and simplifying the expression.

Summary

The problem requires evaluating a limit of an indeterminate form $\frac{0}{0}$. We utilized the Maclaurin series expansions for $\sin^{-1}x$ and $\tan^{-1}x$ to represent the functions around $x=0$. By substituting these series into the limit expression and simplifying, we found that the terms up to $x^3$ were sufficient to determine the limit. After calculating the value of $L$ to be $\frac{1}{6}$, we proceeded to calculate the value of $6L + 1$, which resulted in 2.

The final answer is \boxed{2}. This corresponds to option (D).