Let f(x) = ∫[0 to x] (t + sin(1 - e^t)) dt, x ∈ R. Then, lim(x→0) f(x)/x³ is equal to
(1) -1/6
(2) 2

Q: Why do we have to apply L'Hopital's Rule three times? Is there a faster way?

You are absolutely right to ask! While L'Hopital's Rule is the standard, robust path, you could also use Taylor series expansions for $e^x$ and $\sin(u)$ around $x=0$. However, for JEE Advanced, mastering the repeated application of L'Hopital's Rule is a critical skill because it builds the algebraic stamina needed for complex problems where Taylor series might become messy.

Q: Why does the lower limit of the integral disappear when we differentiate?

Great observation! The Leibniz Rule states that $\frac{d}{dx} \int_{a(x)}^{b(x)} g(t) dt = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$. In our case, the lower limit is the constant $0$. Since the derivative of a constant is $0$, the term $g(0) \cdot 0$ vanishes completely. It is a beautiful simplification that happens often in these types of problems.

Q: I keep making mistakes in the chain rule for $\sin(1 - e^x)$. Any tips?

The key is to treat it like an onion. Peel one layer at a time. The derivative of $\sin(u)$ is $\cos(u)$, and then you must multiply by the derivative of $u$. Here, $u = 1 - e^x$, so its derivative is $-e^x$. Always write out the intermediate steps: $\frac{d}{dx} \sin(1 - e^x) = \cos(1 - e^x) \cdot \frac{d}{dx}(1 - e^x) = \cos(1 - e^x) \cdot (-e^x)$. Meticulousness is your best defense against errors.

Question

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

Options:

(A) $-\frac{1}{6}$

(B) $\frac{2}{3}$

(C) $-\frac{2}{3}$

(D) $\frac{1}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast, complex landscape of functions. You have been given a function defined by an integral:

$$f(x) = \int_{0}^{x} (t + \sin(1 - e^t)) dt$$

Your mission is to find the limit of this function divided by $x^3$ as $x$ approaches zero. This is not just a calculation; it is a dance between integration and differentiation.

### The Indeterminate Trap

Before we rush into the fray, let us pause. In the world of limits, the first step is always to test the waters.

If we substitute $x = 0$ directly into our expression, we find that the integral from $0$ to $0$ is $0$, and the denominator $0^3$ is also $0$. We have arrived at the classic $\frac{0}{0}$ indeterminate form.

This is not a dead end; it is an invitation to use **L'Hopital's Rule**. We need to differentiate the numerator and the denominator until the mystery of the limit is revealed.

### The Leibniz Rule – Our Telescope

To differentiate the numerator, we cannot simply use basic power rules. We are dealing with an integral with a variable upper limit.

This is where the **Leibniz Rule** becomes our most powerful tool. It allows us to peek inside the integral, as the rule states:

$$\frac{d}{dx} \int_{0}^{x} g(t) dt = g(x)$$

Applying this to our numerator, the derivative $f'(x)$ becomes $x + \sin(1 - e^x)$. The denominator, $x^3$, is much friendlier; its derivative is $3x^2$. Now, our limit looks like this:

$$\lim_{x 	o 0} \frac{x + \sin(1 - e^x)}{3x^2}$$

### The Marathon of Differentiation

We test the limit again. Substituting $x = 0$ gives us $0 + \sin(1 - 1) = 0$ in the numerator and $3(0)^2 = 0$ in the denominator.

We are still in $\frac{0}{0}$ territory! We must apply L'Hopital's Rule again. Differentiating the denominator $3x^2$ gives us $6x$.

Now, for the numerator:

$$f''(x) = \frac{d}{dx}(x + \sin(1 - e^x)) = 1 + \cos(1 - e^x) \cdot (-e^x) = 1 - e^x \cos(1 - e^x)$$

Our new limit is:

$$\lim_{x 	o 0} \frac{1 - e^x \cos(1 - e^x)}{6x}$$

### The Final Stretch

One last check. At $x = 0$, the numerator is $1 - e^0 \cos(1 - e^0) = 1 - 1 \cdot 1 = 0$. The denominator is $6(0) = 0$.

We are still at $\frac{0}{0}$! Do not be discouraged; this is the final application of L'Hopital's Rule. Differentiating the denominator $6x$ gives us the constant $6$.

Now, we differentiate the numerator $1 - e^x \cos(1 - e^x)$ one last time using the product rule:

$$\frac{d}{dx}(1 - e^x \cos(1 - e^x)) = - [e^x \cos(1 - e^x) + e^x (-\sin(1 - e^x))(-e^x)]$$

This simplifies to:

$$- [e^x \cos(1 - e^x) + e^{2x} \sin(1 - e^x)]$$

### The Elegant Conclusion

Now, we substitute $x = 0$ into our final expression. The numerator becomes:

$$- [e^0 \cos(1 - e^0) + e^0 \sin(1 - e^0)] = - [1 \cdot 1 + 1 \cdot 0] = -1$$

Our denominator is $6$. Thus, the limit is **$-\frac{1}{6}$**.

We have navigated the complexity, applied the rules with precision, and arrived at the elegant truth. Remember, in JEE Advanced, it is not just about the final answer; it is about the persistence to see the calculation through to the end.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t, x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Identifying the Indeterminate Form

Applying L'Hopital's Rule

The Leibniz Rule for Differentiation

Calculating $f^{'} (x)$

Updating the Limit Expression

Checking the Form Again

Second Application of L'Hopital's Rule

Differentiating the Numerator

The Second Simplified Limit

Third Application of L'Hopital's Rule

Differentiating the Complex Numerator

Substituting $x = 0$

Final Arithmetic Calculation

Conclusion and Key Takeaways

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Indeterminate Trap

The Leibniz Rule – Our Telescope

The Marathon of Differentiation

The Final Stretch

The Elegant Conclusion

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