The value of `lim_(xrarr0)((e^(x)-x-1)(x-sinx)ln(1+x))/(x^(6))` is equal to

Options

A. `(1)/(2)`
B. `(1)/(6)`
C. `(1)/(12)`
D. `(1)/(3)`

Answer

To solve the limit \( \lim_{x \to 0} \frac{(e^x - x - 1)(x - \sin x) \ln(1+x)}{x^6} \), we can follow these steps:

### Step 1: Substitute \( x = 0 \)
First, we substitute \( x = 0 \) into the expression to check if we get an indeterminate form:
\[ e^0 - 0 - 1 = 1 - 0 - 1 = 0, \]
\[ 0 - \sin(0) = 0 - 0 = 0, \]
\[ \ln(1 + 0) = \ln(1) = 0. \]
Thus, the limit evaluates to \( \frac{0 \cdot 0 \cdot 0}{0^6} = \frac{0}{0} \), which is an indeterminate form.

### Step 2: Use Taylor Series Expansions
To resolve the indeterminate form, we can use Taylor series expansions for \( e^x \), \( \sin x \), and \( \ln(1+x) \) around \( x = 0 \).

1. **Expansion of \( e^x \)**:
\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4). \]
Therefore,
\[ e^x - x - 1 = \frac{x^2}{2} + \frac{x^3}{6} + O(x^4). \]

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

3. **Expansion of \( \ln(1+x) \)**:
\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4). \]

### Step 3: Substitute the Expansions into the Limit
Now we substitute these expansions back into the limit:
\[ (e^x - x - 1)(x - \sin x) \ln(1+x) = \left(\frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right)\left(\frac{x^3}{6} + O(x^5)\right)\left(x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)\right). \]

### Step 4: Simplify the Expression
Now we simplify the product:
- The leading term from \( (e^x - x - 1) \) is \( \frac{x^2}{2} \).
- The leading term from \( (x - \sin x) \) is \( \frac{x^3}{6} \).
- The leading term from \( \ln(1+x) \) is \( x \).

Thus,
\[ \left(\frac{x^2}{2}\right)\left(\frac{x^3}{6}\right)(x) = \frac{x^6}{12}. \]

### Step 5: Substitute Back into the Limit
Now we substitute this back into the limit:
\[ \lim_{x \to 0} \frac{\frac{x^6}{12}}{x^6} = \lim_{x \to 0} \frac{1}{12} = \frac{1}{12}. \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{\frac{1}{12}}. \]