_x0(48)/(x^4)_0^x(t^3)/(t^6+1),dt is equal to _______.

Question

$\displaystyle\lim_{x	o0}\frac{48}{x^4}\int_0^x\frac{t^3}{t^6+1}\,dt$ is equal to \_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

This is $\frac00$ form; apply L'Hopital's rule:

$$48\lim_{x\to0}\frac{x^3/(x^6+1)}{4x^3} = 48\cdot\frac14 = 12$$

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