Question:

Step 1: Find the intersection points of the line and the parabola. To find the points where the line $y = kx + k$ intersects the parabola $y = x^2$, we set the expressions for $y$ equal to each other: $$kx + k = x^2$$ Rearrange the equation into a standard quadratic form: $$x^2 - kx - k = 0$$ We use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots, where $a=1$, $b=-k$, and $c=-k$: $$x = \frac{-(-k) \pm \sqrt{(-k)^2 - 4(1)(-k)}}{2(1)}$$ $$x = \frac{k \pm \sqrt{k^2 + 4k}}{2}$$ Let these two intersection points be $x_1$ and $x_2$: $$x_1 = \frac{k - \sqrt{k^2 + 4k}}{2}$$ $$x_2 = \frac{k + \sqrt{k^2 + 4k}}{2}$$ Step 2: Calculate the difference between the intersection points. The difference between the x-coordinates of the intersection points, $(x_2 - x_1)$, represents the width of the region bounded by the curves along the x-axis. $$x_2 - x_1 = \left(\frac{k + \sqrt{k^2 + 4k}}{2}\right) - \left(\frac{k - \sqrt{k^2 + 4k}}{2}\right)$$ $$x_2 - x_1 = \frac{k + \sqrt{k^2 + 4k} - k + \sqrt{k^2 + 4k}}{2}$$ $$x_2 - x_1 = \frac{2\sqrt{k^2 + 4k}}{2}$$ $$x_2 - x_1 = \sqrt{k^2 + 4k}$$ Step 3: Apply the standard formula for the area between a line and a parabola. The area $A_k$ bounded by a line $y = mx+c$ and a parabola $y = ax^2+bx+d$ between their intersection points $x_1$ and $x_2$ is given by the formula $A = \frac{|a|(x_2-x_1)^3}{6}$. In this problem, the parabola is $y = x^2$, so the coefficient $a=1$. The line is $y = kx+k$. The line is above the parabola in the interval $(x_1, x_2)$ as the quadratic $x^2-kx-k$ has a positive leading coefficient, implying the parabola opens upwards and is below the line between its roots. Thus, the area is given by: $$A_k = \frac{(x_2-x_1)^3}{6}$$ Step 4: Substitute the difference of roots into the area formula. Substitute the expression for $(x_2 - x_1)$ from Step 2 into the area formula: $$A_k = \frac{(\sqrt{k^2 + 4k})^3}{6}$$ This can be written as: $$A_k = \frac{(k^2 + 4k)^{3/2}}{6}$$ Step 5: Evaluate the required limit. We need to find the value of $\displaystyle\lim_{k \to \infty} \dfrac{A_k}{k^3}$. Substitute the expression for $A_k$: $$\lim_{k \to \infty} \dfrac{A_k}{k^3} = \lim_{k \to \infty} \dfrac{\frac{(k^2 + 4k)^{3/2}}{6}}{k^3}$$ $$= \lim_{k \to \infty} \dfrac{(k^2 + 4k)^{3/2}}{6k^3}$$ To evaluate this limit, we factor out $k^2$ from the term inside the parenthesis in the numerator: $$= \lim_{k \to \infty} \dfrac{(k^2(1 + \frac{4}{k}))^{3/2}}{6k^3}$$ Apply the exponent to both factors: $$= \lim_{k \to \infty} \dfrac{(k^2)^{3/2} (1 + \frac{4}{k})^{3/2}}{6k^3}$$ Simplify $(k^2)^{3/2}$ to $k^3$: $$= \lim_{k \to \infty} \dfrac{k^3 (1 + \frac{4}{k})^{3/2}}{6k^3}$$ Cancel out $k^3$ from the numerator and denominator: $$= \lim_{k \to \infty} \dfrac{(1 + \frac{4}{k})^{3/2}}{6}$$ As $k \to \infty$, the term $\frac{4}{k}$ approaches $0$: $$= \dfrac{(1 + 0)^{3/2}}{6}$$ $$= \dfrac{1^{3/2}}{6}$$ $$= \dfrac{1}{6}$$ The final answer is $\boxed{\dfrac{1}{6}}$.