The value of ` lim_(x to 0) (tan x - sin s)/(x^(3))` equals

Options

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

Answer

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

### Step-by-Step Solution:

1. **Write the Taylor series expansion for \(\tan x\)**:
\[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7) \]

2. **Write the Taylor series expansion for \(\sin x\)**:
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7) \]

3. **Subtract the series for \(\sin x\) from \(\tan x\)**:
\[ \tan x - \sin x = \left( x + \frac{x^3}{3} + \frac{2x^5}{15} \right) - \left( x - \frac{x^3}{6} + \frac{x^5}{120} \right) \]
Simplifying this gives:
\[ \tan x - \sin x = \left( \frac{x^3}{3} + \frac{x^3}{6} \right) + \left( \frac{2x^5}{15} - \frac{x^5}{120} \right) \]

4. **Combine the \(x^3\) terms**:
\[ \frac{x^3}{3} + \frac{x^3}{6} = \frac{2x^3}{6} + \frac{x^3}{6} = \frac{3x^3}{6} = \frac{x^3}{2} \]

5. **Combine the \(x^5\) terms**:
\[ \frac{2x^5}{15} - \frac{x^5}{120} = \frac{16x^5}{120} - \frac{x^5}{120} = \frac{15x^5}{120} = \frac{x^5}{8} \]

6. **Thus, we have**:
\[ \tan x - \sin x = \frac{x^3}{2} + O(x^5) \]

7. **Now substitute this back into the limit**:
\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{2} + O(x^5)}{x^3} \]

8. **Simplifying the limit**:
\[ = \lim_{x \to 0} \left( \frac{1}{2} + \frac{O(x^5)}{x^3} \right) = \frac{1}{2} + 0 = \frac{1}{2} \]

### Final Answer:
\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2} \]