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Mathematics Question on limits and derivatives

Let f(x)=∫0x(t+sin⁡(1−et)) dt, x∈R.f(x) = \int_{0}^{x} \left( t + \sin\left(1 - e^t\right) \right) \, dt, \, x \in \mathbb{R}. Then lim⁡x→0f(x)x3\lim_{x \to 0} \frac{f(x)}{x^3} is equal to:

A

16\frac{1}{6}

B

−16\frac{-1}{6}

C

−23\frac{-2}{3}

D

23\frac{2}{3}

Answer

−16\frac{-1}{6}

Explanation

Solution

We are tasked with evaluating the following limit:

lim⁡x→0f(x)x3\lim_{x \to 0} \frac{f(x)}{x^3} where f(x)=∫0x(t+sin⁡(1−et)) dtf(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):

f′(x)=x+sin⁡(1−ex)f'(x) = x + \sin(1 - e^x)

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

lim⁡x→0f(x)x3=lim⁡x→0f′(x)3x2\lim_{x \to 0} \frac{f(x)}{x^3} = \lim_{x \to 0} \frac{f'(x)}{3x^2}

This becomes:

lim⁡x→0x+sin⁡(1−ex)3x2\lim_{x \to 0} \frac{x + \sin(1 - e^x)}{3x^2}

We apply L’Hopital’s Rule again:

lim⁡x→01+(−sin⁡(1−ex))⋅(−ex)+cos⁡(1−ex)⋅ex6x\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=0x = 0:

lim⁡x→0−sin⁡(1−ex)⋅ex+cos⁡(1−ex)⋅ex6=−16\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:

−16-\frac{1}{6}