`Lt_(x to 0)(tan^(-1)x - sin^(-1)x)/x^(3)` =

Options

A. `-1`
B. `-1//2`
C. `1//2`
D. 1

Answer

To solve the limit \( \lim_{x \to 0} \frac{\tan^{-1}x - \sin^{-1}x}{x^3} \), we will use the Taylor series expansions for \( \tan^{-1}x \) and \( \sin^{-1}x \) around \( x = 0 \).

### Step-by-Step Solution:

1. **Expand \( \tan^{-1}x \)**:
The Taylor series expansion of \( \tan^{-1}x \) around \( x = 0 \) is:
\[ \tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \]

2. **Expand \( \sin^{-1}x \)**:
The Taylor series expansion of \( \sin^{-1}x \) around \( x = 0 \) is:
\[ \sin^{-1}x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots \]

3. **Subtract the expansions**:
Now, we will subtract the expansions:
\[ \tan^{-1}x - \sin^{-1}x = \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \right) - \left( x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots \right) \]
Simplifying this, we get:
\[ \tan^{-1}x - \sin^{-1}x = -\frac{x^3}{3} - \frac{x^3}{6} + O(x^5) \]
Combining the \( x^3 \) terms:
\[ -\frac{x^3}{3} - \frac{x^3}{6} = -\left(\frac{2}{6} + \frac{1}{6}\right)x^3 = -\frac{3}{6}x^3 = -\frac{1}{2}x^3 \]

4. **Substitute back into the limit**:
Now we substitute this result back into the limit:
\[ \lim_{x \to 0} \frac{\tan^{-1}x - \sin^{-1}x}{x^3} = \lim_{x \to 0} \frac{-\frac{1}{2}x^3 + O(x^5)}{x^3} \]
This simplifies to:
\[ \lim_{x \to 0} \left(-\frac{1}{2} + \frac{O(x^5)}{x^3}\right) \]
As \( x \to 0 \), the term \( \frac{O(x^5)}{x^3} \) approaches 0.

5. **Final result**:
Therefore, we have:
\[ \lim_{x \to 0} \frac{\tan^{-1}x - \sin^{-1}x}{x^3} = -\frac{1}{2} \]

### Conclusion:
The final answer is:
\[ \lim_{x \to 0} \frac{\tan^{-1}x - \sin^{-1}x}{x^3} = -\frac{1}{2} \]