lim (x→π/4) (∫₂ˢᵉᶜ² ˣ f(t) dt) / (x² - π²/16) equals (a) 8/π f(2) (b) 2/π f(2) (c) 2/π f(1/2) (d) 4

Q: Wait, why does the integral $\int_2^{\sec^2 x} f(t) dt$ become zero?

Think of a definite integral as the area under a curve. If the starting point and the ending point are the same, the width of the region is zero. Since our upper limit $\sec^2(\frac{\pi}{4})$ equals $2$, the integral becomes $\int_2^2 f(t) dt$, which is geometrically zero. It's a classic trap, but once you see it, you'll never miss it again!

Q: Why do we ignore the lower limit in the Leibniz Rule?

The full Leibniz Rule is $\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))b'(x) - f(a(x))a'(x)$. In our case, the lower limit is the constant $2$. The derivative of a constant is $0$. So, the term $f(2) \cdot 0$ vanishes. We aren't ignoring it; we are mathematically proving it has no impact on the rate of change!

Q: Is L'Hopital's Rule always the best approach for limits?

It is a powerful weapon, but it is not a magic wand. Always check if the form is $0/0$ or $\infty/\infty$ first. If it is not, L'Hopital will lead you to a wrong answer. Here, the form is perfectly indeterminate, making L'Hopital the most elegant path to the solution.

Question

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

Options:

(A) $\frac{8}{\pi} f(2)$

(B) $\frac{2}{\pi} f(2)$

(C) $\frac{2}{\pi} f(1/2)$

(D) $4 f(2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the following limit:

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

Do not let the notation intimidate you. In JEE Advanced, the most complex-looking problems often hide the most beautiful, simple symmetries. Let us peel back the layers together.

### The Indeterminate Trap

Every limit problem is a story of behavior. We want to know what happens to this fraction as $x$ gets infinitely close to $\frac{\pi}{4}$.

First, let us look at the numerator. As $x 	o \frac{\pi}{4}$, the upper limit of our integral, $\sec^2 x$, approaches $\sec^2(\frac{\pi}{4})$. Since $\sec(\frac{\pi}{4}) = \sqrt{2}$, squaring it gives us exactly $2$.

The numerator becomes $\int_2^2 f(t) dt$. Geometrically, this is the area under a curve from $2$ to $2$. The width is zero, so the area is zero.

Now, look at the denominator: $x^2 - \frac{\pi^2}{16}$. As $x 	o \frac{\pi}{4}$, this becomes $(\frac{\pi}{4})^2 - \frac{\pi^2}{16} = 0$. We have arrived at the classic $0/0$ indeterminate form, which is a green light to use **L'Hopital's Rule**.

### The Weaponry

L'Hopital's Rule states that if we have a $0/0$ form, the limit of the ratio is equal to the limit of the ratio of the derivatives. We need to differentiate the numerator and the denominator separately.

The denominator is straightforward:
$$\frac{d}{dx} \left( x^2 - \frac{\pi^2}{16} ight) = 2x$$

To differentiate the numerator, we use the **Newton-Leibniz Rule**. It states that the derivative of an integral with a variable upper limit $g(x)$ is:
$$\frac{d}{dx} \int_{a}^{g(x)} f(t) dt = f(g(x)) \cdot g'(x)$$

### The Calculus Dance

Our upper limit is $g(x) = \sec^2 x$. Using the chain rule, we find its derivative:
$$\frac{d}{dx}(\sec^2 x) = 2 \sec x \cdot (\sec x 	an x) = 2 \sec^2 x 	an x$$

Assembling the derivative of the numerator, we get:
$$f(\sec^2 x) \cdot 2 \sec^2 x 	an x$$

Since the lower limit is a constant, its derivative is zero. We now substitute these derivatives back into our limit expression:
$$\lim_{x 	o \frac{\pi}{4}} \frac{f(\sec^2 x) \cdot 2 \sec^2 x 	an x}{2x}$$

### Final Calculation

Notice the beauty of the cancellation. The factor of $2$ in the numerator and the $2$ from the denominator derivative cancel out perfectly:
$$\lim_{x 	o \frac{\pi}{4}} \frac{f(\sec^2 x) \sec^2 x 	an x}{x}$$

Now, we substitute $x = \frac{\pi}{4}$. We know $\sec^2(\frac{\pi}{4}) = 2$ and $	an(\frac{\pi}{4}) = 1$. The expression becomes:
$$\frac{f(2) \cdot 2 \cdot 1}{\frac{\pi}{4}}$$

Simplifying this, we arrive at the final result:
$$\frac{2 f(2)}{\frac{\pi}{4}} = \mathbf{\frac{8}{\pi} f(2)}$$

Animated Solution for Mathematics - Definite Integration: $lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

Select Answer:

Visualized Solution

Visualizing the Limit Problem

Evaluating the Numerator Limit

Evaluating the Denominator Limit

Applying L'Hopital's Rule

The Newton-Leibniz Rule

Derivative of the Upper Limit

Derivative of the Numerator

Derivative of the Denominator

The Simplified Limit Expression

Evaluating the Final Limit

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Indeterminate Trap

The Weaponry

The Calculus Dance

Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→π/4​x2−16π2​∫2sec2x​f(t)dt​ equals