Calculus problem #2131

As $P$ tends to $O$ , the lins $OP$ becomes the tangent to the parabola at $O$ , and since $\frac {dy}{dx} = 4x-2$ , this slope is $-2$ . Therefore,

$\displaystyle \lim_{P\rightarrow O} tan \angle AOP = 2$ . Also, $\displaystyle \lim _{P\rightarrow O} tan \angle PAO = 0$ , so that using $tan A + tan B + tan C = tanAtanBtan C$ , we get

$\displaystyle \lim_{P\rightarrow O} tan \angle APO = - 2$ , and hence $sec \angle APO = -\sqrt {5}$ .

Let $\mathbf{a}$ be the vector from $P$ to $O$ , and $\mathbf{b}$ be the vector from $P$ to $A$ . Using the points defined in the problem, we have $\mathbf{a} = \langle -x,-2x(x-1) \rangle,$ and $\mathbf{b} = \langle 1-x, -2x(x-1) \rangle.$ The angle $\angle APO$ between $\mathbf{a}$ and $\mathbf{b}$ is given by $\cos \angle APO = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b|}},$ and $\sec \angle APO = \frac{|\mathbf{a}||\mathbf{b}|}{\mathbf{a} \cdot \mathbf{b}}$ $=\frac{\sqrt{(-x)^2+(-2x(x-1))^2}\sqrt{(1-x)^2+(-2x(x-1))^2}}{\-x(1-x)+(-2x(x-1))^2 }.$ Expanding the numerator and denominator gives $\sec \angle APO = \frac{\sqrt{5x^2-18x^3+45x^4-88x^5-104x^6-64x^7+16x^8}}{-x+5x^2-8x^3+4x^4}.$ So our limit is $\lim_{P \to O} \sec \angle APO$ $=\lim_{x \to 0^+} \frac{\sqrt{5x^2-18x^3+45x^4-88x^5-104x^6-64x^7+16x^8}}{-x+5x^2-8x^3+4x^4}$ $=\lim_{x \to 0^+} \frac{\sqrt{5-18x+45x^2-88x^3-104x^4-64x^5+16x^6}}{-1+5x-8x^2+4x^2} = -\sqrt{5}.$ We were justified in dividing the numerator by $\sqrt{x^2}$ and the denominator by $x$ because they are equal for positive $x$ . Note, however, that if we were interested in the limit from the left, we would have had to divide the denominator by $-x$ , because $\sqrt{x^2}=-x$ for $x<0$ . $-\sqrt{5} \approx \boxed{-2.236}.$