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 problem is to evaluate the limit:
$\lim_{x \rightarrow 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} dt$.
Firstly, let's focus on the integral $\int_0^x \frac{t^3}{t^6+1} dt$. To understand its behavior as $x \rightarrow 0$, consider the function $f(t)=\frac{t^3}{t^6+1}$. Near $t=0$, $f(t) \approx t^3$. This approximation suggests that:
$\int_0^x f(t) dt \approx \int_0^x t^3 dt = \frac{x^4}{4}$.
Substituting this approximation into the original limit, we have:
$\lim_{x \rightarrow 0} \frac{48}{x^4} \cdot \frac{x^4}{4} = \lim_{x \rightarrow 0} 12 = 12$.
Thus, the limit evaluates to 12, which is within the specified range of 12 to 12.