The value of `lim_(xrarr0) (1+sinx-cosx+log(1-x))/(x^3)` , is

Options

A. `1//2`
B. `-1//2`
C. `0`
D. `1`

Answer

To find the limit

\[ \lim_{x \to 0} \frac{1 + \sin x - \cos x + \log(1 - x)}{x^3}, \]

we start by substituting \( x = 0 \):

1. **Evaluate the numerator at \( x = 0 \)**:
- \( 1 + \sin(0) - \cos(0) + \log(1 - 0) = 1 + 0 - 1 + 0 = 0 \).

2. **Evaluate the denominator at \( x = 0 \)**:
- \( x^3 = 0^3 = 0 \).

Since both the numerator and denominator evaluate to 0, we have an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's rule, which states that if we have an indeterminate form, we can take the derivative of the numerator and the derivative of the denominator.

3. **Differentiate the numerator**:
- The derivative of \( 1 \) is \( 0 \).
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( -\cos x \) is \( \sin x \).
- The derivative of \( \log(1 - x) \) is \( \frac{-1}{1 - x} \).

So, the derivative of the numerator is:

\[ 0 + \cos x + \sin x - \frac{1}{1 - x}. \]

4. **Differentiate the denominator**:
- The derivative of \( x^3 \) is \( 3x^2 \).

Now we can rewrite our limit using L'Hôpital's rule:

\[ \lim_{x \to 0} \frac{\cos x + \sin x - \frac{1}{1 - x}}{3x^2}. \]

5. **Evaluate the new limit at \( x = 0 \)**:
- Substitute \( x = 0 \) into the new numerator:
- \( \cos(0) + \sin(0) - \frac{1}{1 - 0} = 1 + 0 - 1 = 0 \).
- The denominator becomes \( 3(0^2) = 0 \).

Again, we have the \( \frac{0}{0} \) form, so we apply L'Hôpital's rule again.

6. **Differentiate the numerator again**:
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( -\frac{1}{1 - x} \) is \( \frac{-(-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2} \).

So, the new numerator is:

\[ -\sin x + \cos x + \frac{1}{(1 - x)^2}. \]

7. **Differentiate the denominator again**:
- The derivative of \( 3x^2 \) is \( 6x \).

Now we rewrite our limit:

\[ \lim_{x \to 0} \frac{-\sin x + \cos x + \frac{1}{(1 - x)^2}}{6x}. \]

8. **Evaluate the new limit at \( x = 0 \)**:
- Substitute \( x = 0 \):
- The numerator becomes \( -\sin(0) + \cos(0) + \frac{1}{(1 - 0)^2} = 0 + 1 + 1 = 2 \).
- The denominator becomes \( 6(0) = 0 \).

Again, we have \( \frac{2}{0} \), which indicates we need to analyze the limit more closely.

9. **Final evaluation**:
- As \( x \to 0 \), the numerator approaches \( 2 \) and the denominator approaches \( 0 \), leading to infinity. However, we need to check the sign of the limit as \( x \) approaches from both sides.

Using the Taylor series expansions around \( x = 0 \):

- \( \sin x \approx x \)
- \( \cos x \approx 1 - \frac{x^2}{2} \)
- \( \log(1 - x) \approx -x \)

Substituting these approximations into the original limit gives us:

\[ \lim_{x \to 0} \frac{1 + x - (1 - \frac{x^2}{2}) - x}{x^3} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{x^3} = \lim_{x \to 0} \frac{1}{2x} = \infty. \]

Thus, the limit diverges.

### Final Answer:
The limit diverges to \( \infty \).