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

The objective is to evaluate the limit:

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

L’Hopital’s Rule is applied. The derivative of \( f(x) \) is:

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

Applying L’Hopital’s Rule to the limit:

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

This simplifies to:

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

L’Hopital’s Rule is applied again:

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

Evaluation at \( x = 0 \):

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

The value of the limit is:

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