Question:

Step 1: Rewrite the limit expression and identify its indeterminate form. Let the given limit be $L$. We can rewrite the expression to clearly identify its form as $x \to 0$. $$L = \lim_{x\to0}\dfrac{48}{x^4}\int_0^x\dfrac{t^3}{t^6+1}\,dt = 48 \lim_{x\to0}\dfrac{\int_0^x\dfrac{t^3}{t^6+1}\,dt}{x^4}$$ As $x \to 0$: The numerator approaches $\int_0^0\dfrac{t^3}{t^6+1}\,dt = 0$. The denominator approaches $0^4 = 0$. Since the limit is of the form $\dfrac{0}{0}$, we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule. According to L'Hôpital's Rule, if $\displaystyle\lim_{x\to a} \dfrac{f(x)}{g(x)}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, then the limit is equal to $\displaystyle\lim_{x\to a} \dfrac{f'(x)}{g'(x)}$. Let $f(x) = \int_0^x\dfrac{t^3}{t^6+1}\,dt$ and $g(x) = x^4$. We need to find the derivatives $f'(x)$ and $g'(x)$. Using the Fundamental Theorem of Calculus, Part I, if $F(x) = \int_a^x h(t)\,dt$, then $F'(x) = h(x)$. So, $f'(x) = \dfrac{x^3}{x^6+1}$. And $g'(x) = \dfrac{d}{dx}(x^4) = 4x^3$. Now, apply L'Hôpital's Rule: $$L = 48 \lim_{x\to0}\dfrac{\dfrac{x^3}{x^6+1}}{4x^3}$$ Step 3: Simplify the expression and evaluate the limit. We can simplify the expression by canceling $x^3$ from the numerator and denominator (since $x \to 0$ but $x \neq 0$): $$L = 48 \lim_{x\to0}\dfrac{x^3}{(x^6+1) \cdot 4x^3}$$ $$L = 48 \lim_{x\to0}\dfrac{1}{4(x^6+1)}$$ Now, substitute $x=0$ into the simplified expression: $$L = 48 \cdot \dfrac{1}{4(0^6+1)}$$ $$L = 48 \cdot \dfrac{1}{4(1)}$$ $$L = 48 \cdot \dfrac{1}{4}$$ $$L = 12$$ Step 4: State the final answer. The value of the limit is $12$. The final answer is $\boxed{12}$.