Question

Let \(R\) be the rectangle joining the midpoints of the sides of the quadrilateral \(Q\) having vertices \((\pm x, 0)\) and \((0, \pm 1)\). Calculate $$ \lim _{x \rightarrow 0^{+}} \frac{\text { perimeter of } R}{\text { perimeter of } Q} $$

Short answer

The limit is \(\frac{1}{2}\).

Step-by-step solution

Calculate Midpoints of Quadrilateral Sides

The quadrilateral \(Q\) has vertices \((x, 0), (-x, 0), (0, 1), (0, -1)\). To form the rectangle \(R\), we find the midpoints of these sides:1. Between \((x, 0)\) and \((0, 1)\): \(\left(\frac{x}{2}, \frac{1}{2}\right)\)2. Between \((x, 0)\) and \((0, -1)\): \(\left(\frac{x}{2}, -\frac{1}{2}\right)\)3. Between \((-x, 0)\) and \((0, 1)\): \(\left(-\frac{x}{2}, \frac{1}{2}\right)\)4. Between \((-x, 0)\) and \((0, -1)\): \(\left(-\frac{x}{2}, -\frac{1}{2}\right)\)

Find the Lengths of Rectangle Sides

Rectangle \(R\) has vertices at \(\left(\frac{x}{2}, \frac{1}{2}\right), \left(\frac{x}{2}, -\frac{1}{2}\right), \left(-\frac{x}{2}, \frac{1}{2}\right), \left(-\frac{x}{2}, -\frac{1}{2}\right)\).- The horizontal side length is the distance between \(\left(\frac{x}{2}, \frac{1}{2}\right)\) and \(\left(-\frac{x}{2}, \frac{1}{2}\right)\), which is \(x\).- The vertical side length is the distance between \(\left(\frac{x}{2}, \frac{1}{2}\right)\) and \(\left(\frac{x}{2}, -\frac{1}{2}\right)\), which is 1.

Calculate the Perimeter of Rectangle R

The perimeter of rectangle \(R\) is calculated using its side lengths: two sides of \(x\) and two sides of 1. The formula for the perimeter of a rectangle is given by\[\text{Perimeter of } R = 2(x + 1).\]

Find the Perimeter of Quadrilateral Q

The quadrilateral \(Q\) is formed by straight calculations between its vertices:- The top horizontal side length, between \((x, 0)\) and \((0, 1)\), is \(\sqrt{x^2 + 1}\).- The bottom horizontal side length, between \(x, 0\) and \((0, -1)\), is also \(\sqrt{x^2 + 1}\).- The left vertical side length, between \((-x, 0)\) and \((0, 1)\), is \(\sqrt{x^2 + 1}\).- The right vertical side length, between \((-x, 0)\) and \((0, -1)\), is also \(\sqrt{x^2 + 1}\).Thus, the perimeter of \(Q\) is\[\text{Perimeter of } Q = 4\sqrt{x^2 + 1}.\]

Limit of Perimeter Ratio as x Approaches 0 from the Right

We are interested in calculating\[\lim _{x \rightarrow 0^{+}} \frac{\text { perimeter of } R}{\text { perimeter of } Q} = \lim _{x \rightarrow 0^{+}} \frac{2(x+1)}{4\sqrt{x^2 + 1}}.\]- As \(x\) approaches 0, the expression simplifies as follows:\[\lim _{x \rightarrow 0^{+}} \frac{2(x+1)}{4\sqrt{x^2 + 1}} = \frac{2 \times 1}{4 \times 1} = \frac{1}{2}.\]Thus, the limit of the ratio of the perimeters is \(\frac{1}{2}\).

Key concepts

Understanding Limits in Calculus

In calculus, limits play a crucial role in understanding how functions behave as inputs approach a particular value. Limits help us predict the behavior of functions near points of interest without necessarily arriving at those points. For example, by evaluating the limit \[ \lim_{x \to 0^+} \frac{2(x+1)}{4\sqrt{x^2 + 1}} \], we are interested in seeing how the ratio of the perimeters changes as the variable \( x \) approaches zero from a positive direction. This is a fascinating concept because although the actual value at \( x = 0 \) may be undefined or different, the trend as we near this point yields significant insights.

By understanding limits, we gain the ability to reason about calculus in a dynamic way. Limits are also foundational for techniques such as differentiation and integration. When learning about limits, remember:

• They are essential for understanding continuous change.
• They provide a way to handle indeterminate forms.
• They form the basis for defining derivatives and integrals.

Calculating Perimeter: Essential Geometry

The perimeter is the total length around a two-dimensional shape. For quadrilaterals like squares, rectangles, and general four-sided figures, calculating the perimeter involves adding up the lengths of all the sides. In our exercise, we have a rectangle and a quadrilateral, which both require perimeter calculation for solving the problem.

To find the perimeter of a rectangle, we use the formula:\[ \text{Perimeter of a rectangle} = 2 ( ext{length} + ext{width} ) \]For our rectangle \( R \) with side lengths \( x \) and 1, this formula gives us the perimeter \( 2(x + 1) \). Contrast this with the perimeter of quadrilateral \( Q \), where each side is calculated using the distance formula:

• The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

This calculation is more intricate for \( Q \) because its vertices do not form perfect horizontal or vertical sides.

What is a Quadrilateral?

A quadrilateral is a polygon with four edges (sides) and four vertices (corners), paving the way for a variety of shapes such as squares, rectangles, trapezoids, and parallelograms. In our problem, quadrilateral \( Q \) is explored to find its perimeter. This can be visualized by its four vertices: \((x, 0), (-x, 0), (0, 1)\), and \((0, -1)\). Being a versatile shape, a quadrilateral can have sides of different lengths and angles that aren't right angles.

Key properties include:

• The sum of interior angles in a quadrilateral is always \( 360^\circ \).
• Quadrilaterals can have parallel sides or none at all, affecting their classification.

Understanding these characteristics helps in applying the correct methods for calculating properties such as perimeter or area and predicting outcomes when limits are involved.

Rectangle: A Special Quadrilateral

A rectangle is a unique type of quadrilateral where all angles are right angles (\(90^\circ\)). Opposite sides of a rectangle are equal in length, making perimeter calculations straightforward. This simplicity stands in contrast to the more general form of quadrilaterals. In our exercise, rectangle \( R \) emerges by joining midpoints of the sides of quadrilateral \( Q \). Because all angles are right angles, its perimeter is simply twice the sum of the lengths of the two adjacent sides.

Understanding rectilinear shapes is crucial since they frequently appear in basic geometry and more complex mathematical problems:

• Rectangles have congruent diagonals that bisect each other.
• They serve as the building blocks for constructions where right angles are essential.

Having clarity on the nature of rectangles helps delineate problem-solving strategies effectively, especially in problems involving transformations, such as calculating perimeters and areas in varied conditions.