Chapter 2.1, Problem 80E

To determine

To find: The value of limit $\underset{P\to O}{\lim }b$ .

Answer to Problem 80E

The value of limit $\underset{P\to O}{\lim }b$ is $\frac{1}{2}$ .

Explanation of Solution

Given information: Let $P\left(a,a^2\right)$ be a point on the parabola $y=x^2$ , $a>0$ . Let $O$ be the origin and $\left(0,b\right)$ the $y$ -intercept of the perpendicular bisector of the line segment $OP$ .

Calculation:

The perpendicular bisector of the line segment $OP$ passes through the midpoint of the line segment $OP$ .

The midpoint of the line segment $OP$ is $\left(\frac{0+a}{2},\frac{0+a^2}{2}\right)=\left(\frac{a}{2},\frac{a^2}{2}\right)$ .

Now use the formula for slope of line segment $OP$ .

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{a^2-0}{a-0}\\ =a\end{array}$

The slope of the line perpendicular to segment $OP$ is $-\frac{1}{m}=-\frac{1}{a}$ .

The equation of the perpendicular bisector to segment $OP$ .

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-\frac{a^2}{2}=-\frac{1}{a}\left(x-\frac{a}{2}\right)\\ y=-\frac{1}{a}x+\left(\frac{1}{2}+\frac{a^2}{2}\right)\end{array}$

Substitute $x=0$ for the $y$ -intercept of the line.

  $\begin{array}{c}y=-\frac{1}{a}x+\left(\frac{1}{2}+\frac{a^2}{2}\right)\\ y=-\frac{1}{a}\times 0+\left(\frac{1}{2}+\frac{a^2}{2}\right)\\ y=\left(\frac{1}{2}+\frac{a^2}{2}\right)\end{array}$

As given the $y$ -intercept is $b$ . So, $b=\frac{1}{2}+\frac{a^2}{2}$ .

As the point $P\left(a,a^2\right)$ approaches to origin the value of $a$ approaches $0$ .

Simplify the limit.

  $\begin{array}{c}\underset{P\to O}{\lim }b=\underset{a\to 0}{\lim }b\\ =\underset{a\to 0}{\lim }\left(\frac{1}{2}+\frac{a^2}{2}\right)\\ =\frac{1}{2}+\frac{0}{2}\\ =\frac{1}{2}\end{array}$

Therefore, the value of limit $\underset{P\to O}{\lim }b$ is $\frac{1}{2}$ .