Question

Let \[ f(x) = \int_{0}^{x} \left( t + \sin\left(1 - e^t\right) \right) \, dt, \, x \in \mathbb{R}. \] Then \[ \lim_{x \to 0} \frac{f(x)}{x^3} \] is equal to:

• A. \( \frac{1}{6} \)
• B. \( \frac{-1}{6} \)
• C. \( \frac{-2}{3} \)
• D. \( \frac{2}{3} \)

Answer 1

The Correct Option is B

We are tasked with evaluating the following limit:

\[ \lim_{x \to 0} \frac{f(x)}{x^3} \] where \( f(x) = \int_{0}^{x} \left(t + \sin \left(1 - e^t\right)\right) \, dt \).

To solve this, we apply L’Hopital’s Rule. First, we compute the derivative of \( f(x) \):

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

Now, applying L’Hopital’s Rule to evaluate the limit:

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

This becomes:

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

We apply L’Hopital’s Rule again:

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

Evaluating this at \( x = 0 \):

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

Thus, the value of the limit is:

\[ -\frac{1}{6} \]

Answer 2

Step 1: Definition of the function \( f(x) \).
\[ f(x) = \int_0^x \left( t + \sin(1 - e^t) \right) dt \] We want to find: \[ \lim_{x \to 0} \frac{f(x)}{x^3} \]
Step 2: Strategy.
Use Taylor series expansions or properties of definite integrals for small \( x \) to find the behavior of \( f(x) \) near 0.

Step 3: Approximate the integrand near \( t=0 \).
- For small \( t \), \( e^t \approx 1 + t + \frac{t^2}{2} + \dots \)
- So, \( 1 - e^t \approx 1 - (1 + t + \frac{t^2}{2}) = -t - \frac{t^2}{2} + \dots \)
- Now, \( \sin(1 - e^t) \approx \sin(-t - \frac{t^2}{2}) \). Since sine is odd, \( \sin(-\theta) = -\sin(\theta) \), this becomes:
\[ -\sin\left(t + \frac{t^2}{2}\right) \approx -(t + \frac{t^2}{2}) + \dots = -t - \frac{t^2}{2} + \dots \]
- Thus, the integrand near 0 becomes:
\[ t + \sin(1 - e^t) \approx t - t - \frac{t^2}{2} = -\frac{t^2}{2} + \dots \]
Step 4: Integrate the approximation.
\[ f(x) \approx \int_0^x \left( -\frac{t^2}{2} \right) dt = -\frac{1}{2} \int_0^x t^2 dt = -\frac{1}{2} \times \frac{x^3}{3} = -\frac{x^3}{6} \]
Step 5: Compute the limit.
\[ \lim_{x \to 0} \frac{f(x)}{x^3} = \lim_{x \to 0} \frac{-\frac{x^3}{6}}{x^3} = -\frac{1}{6} \]
Final Answer:
\[ \boxed{-\frac{1}{6}} \]