Question

$\underset{x\to 0}{\lim }\frac{48}{x^4}\underset{0}{\overset{x}{\int }}\frac{t^3}{t^6+1}dt$ is equal to

Answer 1

Answer: 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
$48\underset{x\to 0}{\lim }\frac{x^3}{x^6+1}\times \frac{1}{4x^3}$
$=12$