x 0 48x^4 0^x t^3t^6 + 1 dt is equal to. — Solution | JEE Mathematics

NVAMediumJEE 2023Limits

$\lim_{x \to 0} \frac{48}{x^4} \int_{0}^{x} \frac{t^3}{t^6 + 1} \, dt$ is equal to _____.

Answer

Correct answer:12

Step-by-step solution

Standard Method

Given: The question asks for

\lim_{x \to 0} \frac{48}{x^4} \int_{0}^{x} \frac{t^3}{t^6 + 1} \, dt

Find: The numerical value of this limit.

The solution is for a different question about the area bounded by curves, not for this limit problem. It concludes with $25$ , but that result is unrelated to the given question.

Therefore, the correct answer cannot be reliably derived from the solution, and the answer key also conflicts with it. Hence the answer is AMBIGUOUS.

Common mistakes

Using the unrelated area-between-curves solution to answer this limit question. That solution does not correspond to the given integrand, so its final value cannot be used here. Always verify that the solution matches the question statement.
Substituting $x = 0$ directly into
$\frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} \, dt$
and stopping at an indeterminate form without further analysis. For such limits, expand the integrand near $t=0$ or apply an appropriate theorem carefully.

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Answer

Step-by-step solution

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