Let f : R → R be a continuous function. Then lim(x→π/4) [(π/4) ∫(2 to sec²x) f(x) dx] / (x² - π²/16)

Q: Wait, why do we use L'Hopital's Rule here? Can't we just solve the integral?

That is a great question! We don't know the explicit form of the function $f(t)$, so we cannot perform the integration directly. L'Hopital's Rule is our bridge to bypass the unknown integral by focusing on its rate of change.

Q: Why does the lower limit of the integral (which is 2) disappear when we differentiate?

Remember the Fundamental Theorem of Calculus: the derivative of a constant is zero. Since the lower limit is a fixed constant, it contributes nothing to the rate of change of the integral as $x$ varies.

Q: I keep messing up the chain rule for $\sec^2(x)$. Any tips?

Think of $\sec^2(x)$ as $(\sec x)^2$. When you differentiate, bring the power down first to get $2(\sec x)$, then multiply by the derivative of the inside, which is $\sec x \tan x$. It becomes $2 \sec^2 x \tan x$!

Q: Is it always safe to use L'Hopital's Rule for any limit?

Only if the limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Always check the form first, just like we did in the first two steps of our solution.

Q: Why is the final answer $2f(2)$ and not just $f(2)$?

Don't forget the constants! We had a $\frac{\pi}{4}$ in the numerator and a $2x$ in the denominator. When you plug in $x = \frac{\pi}{4}$, those constants don't just vanish; they combine to give you that factor of 2.

Question

Let $f : R 	o R$ be a continuous function. Then $\lim_{x 	o \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_2^{\sec^2 x} f(x) dx}{x^2 - \frac{\pi^2}{16}}$ is equal to :

Options:

(A) $f(2)$

(B) $2f(2)$

(C) $2f(\sqrt{2})$

(D) $4f(2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to tackle a problem that looks intimidating at first glance—a limit involving an integral with a variable upper bound.

It is a classic JEE Advanced challenge, but I want you to see it not as a hurdle, but as a beautiful dance between functions and their rates of change.

### The Indeterminate Encounter

Imagine you are standing before this expression:

$$\lim_{x 	o \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_2^{\sec^2 x} f(t) dt}{x^2 - \frac{\pi^2}{16}}$$

Your first instinct might be to panic. How do we integrate a function $f(t)$ when we don't even know what it is?

But wait—take a deep breath. In calculus, whenever you see a limit, your first duty is to test the waters. If we substitute $x = \frac{\pi}{4}$, the upper limit of our integral becomes $\sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2$.

Suddenly, the integral becomes $\int_2^2 f(t) dt$, which is zero. The denominator also vanishes to zero. We have found ourselves in the land of the $\frac{0}{0}$ indeterminate form. This is not a dead end; it is an invitation to use **L'Hopital's Rule**.

### The Power of Leibniz

To apply L'Hopital's Rule, we must differentiate the numerator and the denominator separately. The denominator is easy: $\frac{d}{dx}(x^2 - \frac{\pi^2}{16}) = 2x$.

But what about that integral? This is where the **Newton-Leibniz Formula** comes to our rescue. It tells us that the derivative of an integral with a variable upper limit $g(x)$ is simply the function evaluated at that limit, multiplied by the derivative of the limit itself:

$$\frac{d}{dx} \int_{a}^{g(x)} f(t) dt = f(g(x)) \cdot g'(x)$$

It is elegant, isn't it? We don't need to know the antiderivative of $f(t)$. We only need to know how the boundary of the integral moves.

### The Chain Rule Symphony

Now, let us focus on the upper limit $g(x) = \sec^2 x$. We need its derivative. Using the chain rule, we differentiate the outer power first, then the inner trigonometric function:

$$\frac{d}{dx}(\sec^2 x) = 2 \sec x \cdot \frac{d}{dx}(\sec x) = 2 \sec x \cdot (\sec x 	an x) = 2 \sec^2 x 	an x$$

Now, let us assemble our pieces. After applying the derivative to the numerator, we have:

$$\frac{\pi}{4} \cdot f(\sec^2 x) \cdot 2 \sec^2 x 	an x$$

### The Final Convergence

We are almost there. We have our new limit expression:

$$\lim_{x 	o \frac{\pi}{4}} \frac{\frac{\pi}{4} \left[ f(\sec^2 x) \cdot 2 \sec^2 x 	an x ight]}{2x}$$

Now, we evaluate this at $x = \frac{\pi}{4}$. We know that $\sec^2(\frac{\pi}{4}) = 2$ and $	an(\frac{\pi}{4}) = 1$. Substituting these values in, the numerator becomes:

$$\frac{\pi}{4} \cdot f(2) \cdot (2 \cdot 2 \cdot 1) = \pi f(2)$$

And the denominator becomes $2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$. When we divide the two, the $\pi$ terms cancel out beautifully, leaving us with the final result:

**Result = $2f(2)$**

### Reflection

Look at what we have achieved. We started with a terrifying integral and, through the systematic application of the Leibniz Rule and L'Hopital's Rule, we distilled it down to a simple, elegant result.

This is the essence of JEE Advanced mathematics: it is not about memorizing formulas, but about understanding the underlying structure of the problem. You have the tools; you have the logic. Keep practicing, keep questioning, and most importantly, keep enjoying the process.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a continuous function. Then $lim_{x \to \frac{π}{4}} \frac{\frac{π}{4} \int _{2}^{s e c^{2} x} f ( x ) d x}{x ^{2} - \frac{π ^{2}}{16}}$ is equal to :

Select Answer:

Visualized Solution

The Given Limit

Evaluating the Numerator at $x = \frac{π}{4}$

Evaluating the Denominator at $x = \frac{π}{4}$

Applying L'Hopital's Rule

The Newton-Leibniz Formula

Applying Leibniz Rule

Differentiating the Upper Limit

Derivative of the Denominator

The New Limit Expression

Evaluating at $x = \frac{π}{4}$

Simplifying the Numerator

Final Simplification

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

Analyzing the Setup

The Indeterminate Encounter

The Power of Leibniz

The Chain Rule Symphony

The Final Convergence

Reflection

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