Question

$\displaystyle\lim _{x \rightarrow 0} \frac{48}{x^4} \int_0^x \frac{ t ^3}{ t ^6+1} d t$ is equal to

Answer 1

The correct answer is 12.
48x→0lim​x40∫x​t6+1t3​dt​(00​)
Applying L' Hospitals Rule
48x→0lim​x6+1x3​×4x31​
=12