Let `f(x) = int_0^x(t + sin (1 – e^t))dt, x in R`. Then, `lim_(xrarr0) frac{f(x)}{x^3}` is equal to

Options

A. `-frac{1}{6}`
B. `2/3`
C. `-frac{2}{3}`
D. `1/6`

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{f(x)}{x^3} \]

where

\[ f(x) = \int_0^x (t + \sin(1 - e^t)) dt. \]

### Step 1: Evaluate \( f(0) \)

First, we find \( f(0) \):

\[ f(0) = \int_0^0 (t + \sin(1 - e^t)) dt = 0. \]

### Step 2: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \( x \to 0 \), we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{f(x)}{x^3} = \lim_{x \to 0} \frac{f'(x)}{3x^2}. \]

### Step 3: Differentiate \( f(x) \)

Using the Fundamental Theorem of Calculus, we differentiate \( f(x) \):

\[ f'(x) = x + \sin(1 - e^x). \]

### Step 4: Substitute into the limit

Now we substitute \( f'(x) \) into the limit:

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

### Step 5: Evaluate the limit

As \( x \to 0 \), we need to evaluate the numerator:

1. \( x \to 0 \)
2. For \( \sin(1 - e^x) \):
- As \( x \to 0 \), \( e^x \to 1 \) so \( 1 - e^x \to 0 \).
- Therefore, \( \sin(1 - e^x) \to \sin(0) = 0 \).

Thus, the numerator approaches \( 0 + 0 = 0 \) as \( x \to 0 \), leading to another \( \frac{0}{0} \) form. We apply L'Hôpital's Rule again:

\[ \lim_{x \to 0} \frac{f'(x)}{3x^2} = \lim_{x \to 0} \frac{1 + \cos(1 - e^x)(-e^x)}{6x}. \]

### Step 6: Differentiate again

Now we differentiate the numerator:

1. The derivative of \( 1 \) is \( 0 \).
2. For \( \cos(1 - e^x)(-e^x) \), we apply the product rule:
- The derivative is \( -e^x \cos(1 - e^x) + \sin(1 - e^x)e^x \).

Thus, we have:

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

### Step 7: Substitute \( x = 0 \)

Now substituting \( x = 0 \):

1. \( e^0 = 1 \)
2. \( \cos(1 - e^0) = \cos(0) = 1 \)
3. \( \sin(1 - e^0) = \sin(0) = 0 \)

Thus, the limit becomes:

\[ \frac{-1 \cdot 1 + 0}{6} = \frac{-1}{6}. \]

### Final Result

Therefore, the limit is:

\[ \lim_{x \to 0} \frac{f(x)}{x^3} = -\frac{1}{6}. \]