Question:

Step 1: Define coordinates and radius. Let the center of the circle $O$ be at the origin $(0,0)$. The radius is $r=2$. The points $A$ and $B$ on the circle are chosen symmetrically with respect to the x-axis. $$ A = (2\cos(x/2), 2\sin(x/2)) $$ $$ B = (2\cos(x/2), -2\sin(x/2)) $$ Step 2: Calculate Area($\triangle OAB$). The base of $\triangle OAB$ is the segment $AB$. The length of $AB$ is $2 \cdot (2\sin(x/2)) = 4\sin(x/2)$. The height of $\triangle OAB$ from $O$ to $AB$ is the x-coordinate of $A$ (or $B$), which is $2\cos(x/2)$. $$ \text{Area}(\triangle OAB) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot (4\sin(x/2)) \cdot (2\cos(x/2)) $$ Using the identity $2\sin\theta\cos\theta = \sin(2\theta)$: $$ \text{Area}(\triangle OAB) = 4\sin(x/2)\cos(x/2) = 2\sin x $$ Step 3: Determine the coordinates of point $D$. The tangent at $A$ is perpendicular to $OA$. The tangent at $B$ is perpendicular to $OB$. The point $D$ is the intersection of these tangents. By symmetry, $D$ lies on the x-axis. The equation of the tangent at a point $(x_0, y_0)$ on a circle $X^2+Y^2=r^2$ is $x_0 X + y_0 Y = r^2$. For point $A(2\cos(x/2), 2\sin(x/2))$ and $r=2$: $$ 2\cos(x/2) X + 2\sin(x/2) Y = 4 $$ Since $D=(X_D, 0)$, substitute $Y=0$: $$ 2\cos(x/2) X_D = 4 \implies X_D = \frac{2}{\cos(x/2)} $$ Thus, $D = \left(\frac{2}{\cos(x/2)}, 0\right)$. Step 4: Determine the coordinates of point $C$. Point $C$ is the intersection of line $OA$ and line $BD$. Line $OA$ passes through $O(0,0)$ and $A(2\cos(x/2), 2\sin(x/2))$. Its equation is $Y = \tan(x/2) X$. Line $BD$ passes through $B(2\cos(x/2), -2\sin(x/2))$ and $D(2/\cos(x/2), 0)$. The slope of line $BD$ is: $$ m_{BD} = \frac{0 - (-2\sin(x/2))}{2/\cos(x/2) - 2\cos(x/2)} = \frac{2\sin(x/2)}{2\left(\frac{1-\cos^2(x/2)}{\cos(x/2)}\right)} = \frac{\sin(x/2)\cos(x/2)}{\sin^2(x/2)} = \cot(x/2) $$ The equation of line $BD$ is $Y - 0 = \cot(x/2) \left(X - \frac{2}{\cos(x/2)}\right)$. To find $C(X_C, Y_C)$, set the $Y$ values equal: $$ \tan(x/2) X_C = \cot(x/2) \left(X_C - \frac{2}{\cos(x/2)}\right) $$ $$ \frac{\sin(x/2)}{\cos(x/2)} X_C = \frac{\cos(x/2)}{\sin(x/2)} X_C - \frac{2\cos(x/2)}{\sin(x/2)\cos(x/2)} $$ $$ X_C \left(\frac{\sin(x/2)}{\cos(x/2)} - \frac{\cos(x/2)}{\sin(x/2)}\right) = -\frac{2}{\sin(x/2)} $$ $$ X_C \left(\frac{\sin^2(x/2) - \cos^2(x/2)}{\sin(x/2)\cos(x/2)}\right) = -\frac{2}{\sin(x/2)} $$ Using $\cos x = \cos^2(x/2) - \sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$: $$ X_C \left(\frac{-\cos x}{\frac{1}{2}\sin x}\right) = -\frac{2}{\sin(x/2)} $$ $$ X_C \left(\frac{-2\cos x}{\sin x}\right) = -\frac{2}{\sin(x/2)} $$ $$ X_C = \frac{\sin x}{\cos x \sin(x/2)} = \frac{2\sin(x/2)\cos(x/2)}{\cos x \sin(x/2)} = \frac{2\cos(x/2)}{\cos x} $$ Then $Y_C = \tan(x/2) X_C = \frac{\sin(x/2)}{\cos(x/2)} \cdot \frac{2\cos(x/2)}{\cos x} = \frac{2\sin(x/2)}{\cos x}$. So, $C = \left(\frac{2\cos(x/2)}{\cos x}, \frac{2\sin(x/2)}{\cos x}\right)$. Step 5: Calculate Area($\triangle OBC$) and its limit. The area of $\triangle OBC$ can be calculated using the formula $\frac{1}{2} |x_B y_C - x_C y_B|$. $$ x_B y_C = 2\cos(x/2) \cdot \frac{2\sin(x/2)}{\cos x} = \frac{4\sin(x/2)\cos(x/2)}{\cos x} = \frac{2\sin x}{\cos x} $$ $$ x_C y_B = \frac{2\cos(x/2)}{\cos x} \cdot (-2\sin(x/2)) = -\frac{4\sin(x/2)\cos(x/2)}{\cos x} = -\frac{2\sin x}{\cos x} $$ $$ \text{Area}(\triangle OBC) = \frac{1}{2} \left| \frac{2\sin x}{\cos x} - \left(-\frac{2\sin x}{\cos x}\right) \right| = \frac{1}{2} \left| \frac{4\sin x}{\cos x} \right| = \frac{2\sin x}{\cos x} $$ Now, compute the limit: $$ \lim_{x \to 0} \frac{\text{Area}(\triangle OBC)}{\text{Area}(\triangle OAB)} = \lim_{x \to 0} \frac{\frac{2\sin x}{\cos x}}{2\sin x} = \lim_{x \to 0} \frac{1}{\cos x} = \frac{1}{1} = 1 $$ Step 6: Calculate Area($\triangle ADB$). The base of $\triangle ADB$ is $AB = 4\sin(x/2)$. The height of $\triangle ADB$ from $D$ to the line containing $AB$ is the horizontal distance between $D$'s x-coordinate and the x-coordinate of $A$ (or $B$). The line containing $AB$ is $X=2\cos(x/2)$. $$ \text{Height} = \left|\frac{2}{\cos(x/2)} - 2\cos(x/2)\right| = 2\left(\frac{1-\cos^2(x/2)}{\cos(x/2)}\right) = 2\frac{\sin^2(x/2)}{\cos(x/2)} = 2\sin(x/2)\tan(x/2) $$ $$ \text{Area}(\triangle ADB) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot (4\sin(x/2)) \cdot (2\sin(x/2)\tan(x/2)) $$ $$ \text{Area}(\triangle ADB) = 4\sin^2(x/2)\tan(x/2) $$ Step 7: Compute the limit of $\frac{\text{Area}(\triangle ADB)}{(\text{Area}(\triangle OAB))^3}$. Substitute the expressions for the areas: $$ \frac{\text{Area}(\triangle ADB)}{(\text{Area}(\triangle OAB))^3} = \frac{4\sin^2(x/2)\tan(x/2)}{(2\sin x)^3} $$ Use the identities $\sin x = 2\sin(x/2)\cos(x/2)$ and $\tan(x/2) = \frac{\sin(x/2)}{\cos(x/2)}$: $$ = \frac{4\sin^2(x/2) \cdot \frac{\sin(x/2)}{\cos(x/2)}}{8(2\sin(x/2)\cos(x/2))^3} $$ $$ = \frac{\frac{4\sin^3(x/2)}{\cos(x/2)}}{8 \cdot 8\sin^3(x/2)\cos^3(x/2)} $$ $$ = \frac{4\sin^3(x/2)}{64\sin^3(x/2)\cos^4(x/2)} $$ For $x \to 0$, $\sin(x/2) \neq 0$, so $\sin^3(x/2)$ can be cancelled: $$ = \frac{4}{64\cos^4(x/2)} = \frac{1}{16\cos^4(x/2)} $$ As $x \to 0$, $\cos(x/2) \to \cos(0) = 1$. $$ \lim_{x \to 0} \frac{\text{Area}(\triangle ADB)}{(\text{Area}(\triangle OAB))^3} = \frac{1}{16 \cdot 1^4} = \frac{1}{16} $$