Let f : [-π/2, π/2] → R be a differentiable function such that f(0) = 1/2. If lim x→0 (x ∫[0 to x] f

Q: Wait, why can I just split the limit into two parts? Is that always allowed?

Great question! You can split a limit into a product of two limits, $\lim (A \cdot B) = (\lim A) \cdot (\lim B)$, provided that both individual limits exist and are finite. In this problem, both parts converge to finite values, so the splitting is mathematically sound and makes our lives much easier!

Q: Why did we use L'Hopital's Rule instead of Taylor series expansion?

You absolutely could use Taylor series! Expanding $e^{x^2}$ as $1 + x^2 + \dots$ would also lead you to the same result. However, L'Hopital's Rule is often faster when you have a simple integral in the numerator, as the Fundamental Theorem of Calculus makes the derivative of the integral trivial.

Q: Does the fact that $f$ is differentiable matter, or is continuity enough?

For the final step where we say $\lim_{x \to 0} f(x) = f(0)$, continuity is sufficient. Since the problem states $f$ is differentiable, it is automatically continuous, so we are perfectly safe to evaluate the limit by direct substitution.

Q: I got confused by the $x$ in the numerator. Where did it go?

We strategically rearranged the expression to $\frac{\int_0^x f(t)dt}{x} \cdot \frac{x^2}{e^{x^2}-1}$. The $x$ from the original numerator was moved to the denominator of the integral term to create the standard form for the derivative of an integral.

Q: Is there any trap with the interval $[-\pi/2, \pi/2]$?

In this specific problem, the interval defines the domain where $f$ is well-behaved. Since we are taking the limit as $x \to 0$, which is well within the interior of this interval, we don't need to worry about boundary conditions. It's just there to ensure the function is defined near zero!

Question

Let $f : [-\frac{\pi}{2}, \frac{\pi}{2}] 	o \mathbb{R}$ be a differentiable function such that $f(0) = \frac{1}{2}$. If $\lim_{x 	o 0} \frac{x \int_0^x f(t)dt}{e^{x^2}-1} = \alpha$, then $8\alpha^2$ is equal to :

Options:

(A) 16

(B) 2

(C) 1

(D) 4

Sigma Percentile · Accepted Answer

### The Dance of Limits: Unveiling the Hidden Structure

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that at first glance looks like a tangled mess of calculus, but beneath the surface, it is a beautiful, elegant dance of limits.

When you see an expression like $\lim_{x 	o 0} \frac{x \int_0^x f(t)dt}{e^{x^2}-1} = \alpha$, your first instinct might be panic. But take a deep breath. In the world of JEE Advanced, complexity is often just a mask for simplicity.

### Phase 1: The Diagnostic Check

Before we touch any heavy machinery, we must diagnose the patient. As $x$ approaches $0$, the numerator becomes $0 \cdot \int_0^0 f(t)dt$, which is $0$.

The denominator, $e^{x^2} - 1$, becomes $e^0 - 1 = 0$. We have a classic $\frac{0}{0}$ indeterminate form. This is our green light! It tells us that there is a finite value waiting to be uncovered, and we have the tools to find it.

### Phase 2: Strategic Rearrangement

Instead of blindly applying L'Hopital's Rule to the entire expression—which would get messy very quickly—let's be architects. We want to isolate the parts we recognize.

We know that $\lim_{u 	o 0} \frac{e^u - 1}{u} = 1$. Our denominator has $e^{x^2} - 1$. If we divide it by $x^2$, we get a standard limit. Let's rewrite our expression:

$$\alpha = \lim_{x 	o 0} \left( \frac{\int_0^x f(t)dt}{x} \cdot \frac{x^2}{e^{x^2}-1} ight)$$

By pulling the $x^2$ out of the numerator and placing it under the exponential term, we have created a perfect environment. The second part, $\lim_{x 	o 0} \frac{x^2}{e^{x^2}-1}$, is simply the reciprocal of the standard limit, which evaluates to $1$.

Now, the problem has simplified significantly to:

$$\alpha = \lim_{x 	o 0} \frac{\int_0^x f(t)dt}{x}$$

### Phase 3: The Power of the Fundamental Theorem

We are left with $\frac{\int_0^x f(t)dt}{x}$. Again, as $x 	o 0$, this is $\frac{0}{0}$.

This is where the Fundamental Theorem of Calculus shines. When you see an integral with a variable limit, think of it as a function waiting to be differentiated. Applying L'Hopital's Rule, we differentiate the numerator and the denominator with respect to $x$:

$$\alpha = \lim_{x 	o 0} \frac{\frac{d}{dx} \left( \int_0^x f(t)dt ight)}{\frac{d}{dx}(x)}$$

The derivative of the integral $\int_0^x f(t)dt$ is simply $f(x)$, and the derivative of $x$ is $1$. Thus, we arrive at:

$$\alpha = \lim_{x 	o 0} f(x)$$

### Phase 4: The Final Revelation

Since the problem guarantees that $f$ is differentiable, it is inherently continuous at $x=0$. Therefore, $\lim_{x 	o 0} f(x) = f(0)$.

We were given $f(0) = \frac{1}{2}$ right at the start! So, $\alpha = \frac{1}{2}$. Finally, the question asks for $8\alpha^2$.

Substituting our value:

$$8\alpha^2 = 8 \cdot \left(\frac{1}{2}ight)^2 = 8 \cdot \frac{1}{4} = 2$$

And there it is. Through careful rearrangement and the application of fundamental principles, the complexity vanished. Remember, in JEE Advanced, the goal isn't to calculate harder; it's to see clearer. Keep practicing, keep questioning, and most importantly, keep enjoying the elegance of the math!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ is equal to :

Select Answer:

Visualized Solution

The Limit Expression

Identifying the Indeterminate Form

Strategic Rearrangement

The Exponential Limit

Simplifying the Expression

Checking the New Form

Applying L'Hopital's Rule

Fundamental Theorem of Calculus

Evaluating the Limit

Finding $α$

Final Calculation Setup

The Final Answer

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

The Dance of Limits

Phase 1

Phase 2

Phase 3

Phase 4

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