lim x 0 48 x 4 0 x t 3 t 6 + 1 d t is equal to ______. [2023]

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Solution

Q.
$\lim_{x \to 0} \frac{48}{x^{4}} \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t is equal to$ ______. [2023]

Ans.
(12)

Let $I = \lim_{x \to 0} \frac{48 \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t}{x^{4}}$ $(\frac{0}{0} form)$

$I = \lim_{x \to 0} \frac{48 (\frac{x^{3}}{x^{6} + 1})}{4 x^{3}} = \lim_{x \to 0} \frac{48 \cdot x^{3}}{x^{6} + 1} \times \frac{1}{4 x^{3}} = \lim_{x \to 0} \frac{12}{x^{6} + 1}$

$I = 12$

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