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Q. $\displaystyle\lim _{x \rightarrow 0} \frac{48}{x^4} \int\limits_0^x \frac{ t ^3}{ t ^6+1} d t$ is equal to

JEE MainJEE Main 2023Limits and Derivatives

Solution:

$48 \displaystyle\lim _{x \rightarrow 0} \frac{\int\limits_0^x \frac{t^3}{t^6+1} d t}{x^4}\left(\frac{0}{0}\right)$
Applying L' Hospitals Rule
$ 48 \displaystyle\lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3} $
$= 12$