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

Correct Answer: 12

The correct answer is 12.
$48\underset{x\to 0}{\lim }\frac{\underset{0}{\overset{x}{\int }}\frac{t^3}{t^6+1}dt}{x^4}\left(\frac{0}{0}\right)$
Applying L' Hospitals Rule
$48\underset{x\to 0}{\lim }\frac{x^3}{x^6+1}\times \frac{1}{4x^3}$
$=12$