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

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Limits

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

is equal to ____.

Answer: 12

The limit has the indeterminate form

\frac{0}{0}

, so differentiating numerator and denominator (with

\frac{d}{d x} \int_{0}^{x} \frac{t ^{3}}{t ^{6} + 1} d t = \frac{x ^{3}}{x ^{6} + 1}

) gives

x \to 0 lim \frac{48}{x ^{4}} \int_{0}^{x} \frac{t ^{3}}{t ^{6} + 1} d t = 48 x \to 0 lim \frac{x ^{3} / ( x ^{6} + 1 )}{4 x ^{3}} = 48 x \to 0 lim \frac{1}{4 ( x ^{6} + 1 )} = \frac{48}{4} = 12.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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