$$\lim_{x \to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} dt$$ is equal to

We need to evaluate $$\lim_{x \to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} dt$$. As $$x \to 0$$, both $$\int_0^x \frac{t^3}{t^6+1}dt \to 0$$ and $$x^4 \to 0$$, so the expression is of the indeterminate form $$\frac{0}{0}$$.

To resolve this, we rewrite the limit as $$\lim_{x \to 0} \frac{48 \int_0^x \frac{t^3}{t^6+1}dt}{x^4}$$ and apply L'Hopital's rule, differentiating numerator and denominator with respect to $$x$$. By the Fundamental Theorem of Calculus, $$\frac{d}{dx}\int_0^x f(t)dt = f(x)$$, hence the limit becomes $$\lim_{x \to 0} \frac{48 \cdot \frac{x^3}{x^6+1}}{4x^3}$$.

Finally, simplifying inside the limit yields $$\lim_{x \to 0} \frac{48}{4(x^6+1)} = \frac{48}{4(0+1)} = \frac{48}{4} = 12$$, so the value of the original limit is 12.