Evaluating Indeterminate Limits Involving Definite Integrals via Series Expansion and L'Hôpital's Rule

Q: Why can we replace the integrand variable t directly with x when differentiating?

By the fundamental theorem of calculus (Leibniz rule), d/dx integral from 0 to x of f(t) dt = f(x) · d/dx(x) - f(0) · d/dx(0) = f(x).

lim as x→ 0 of 48/(x^4) integral from 0 to x of (t^3)/(t^6+1) dt is equal to

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

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Why can we replace the integrand variable t directly with x when differentiating?

By the fundamental theorem of calculus (Leibniz rule), $\frac{d}{d x} \int_{0}^{x} f (t) d t = f (x) \cdot \frac{d}{d x} (x) - f (0) \cdot \frac{d}{d x} (0) = f (x)$ .

Answer

The limit evaluates to 12.

Why each option works or fails

Step-by-step solution

Given: Expression: $L = \lim_{x \to 0} \frac{48}{x^{4}} \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t$ .
Goal: Evaluate the limit $L$ as $x \to 0$ .
Approach: As $x \to 0$ , $\int_{0}^{0} \frac{t^{3}}{t^{6} + 1} d t = 0$ and $x^{4} \to 0$ , giving a $\frac{0}{0}$ indeterminate form. Apply L'Hôpital's Rule alongside the Leibniz Integral Rule for differentiation of the numerator.
Execute: Differentiate numerator and denominator with respect to $x$ : Numerator derivative: $\frac{d}{d x} (48 \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t) = 48 \cdot \frac{x^{3}}{x^{6} + 1}$ . Denominator derivative: $\frac{d}{d x} (x^{4}) = 4 x^{3}$ . Thus, $L = \lim_{x \to 0} \frac{48 \cdot \frac{x^{3}}{x^{6} + 1}}{4 x^{3}} = \lim_{x \to 0} \frac{48}{4 (x^{6} + 1)}$ .
Verify: Near $t = 0$ , $\frac{t^{3}}{t^{6} + 1} \approx t^{3}$ , so $\int_{0}^{x} t^{3} d t = \frac{x^{4}}{4}$ . Thus, $\frac{48}{x^{4}} (\frac{x^{4}}{4}) = 12$ . The result is consistent.

Shortcut:

When to use it: When the integrand has a simple Maclaurin expansion near the expansion point 0.

Quick kill: Since $t \to 0$ , expand $\frac{t^{3}}{t^{6} + 1} = t^{3} (1 - t^{6} + \dots) = t^{3} + O (t^{9})$ . Integrating gives $\int_{0}^{x} t^{3} d t = \frac{x^{4}}{4}$ , hence $\lim_{x \to 0} 48 \cdot \frac{x^{4} / 4}{x^{4}} = 12$ .

Evaluating Indeterminate Limits Involving Definite Integrals via Series Expansion and L'Hôpital's Rule

What feels right?

The limit evaluates to 12.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

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