Limit Problem: ratio of quadrilaterals' perimeters

[reehlgirl1986]

The problem states:

Let "R" be the rectangle joining the midpoints of the side of the quadrilateral "Q" having the vertices (±x, 0) and (0, ±1). Calculate the following:

. . . . .lim[x->0⁺] (perimeter of R)/(perimeter of Q)

What I would like to know is how to set up the problem. What does it look like graphed? I think it would then help me get started.

 

[soroban]

Re: Limit Problem!

Hello, reehlgirl1986!

Did you make a sketch?

:et \(\displaystyle R\) be the rectangle joining the midpoints of the side of the quadrilateral \(\displaystyle Q\)
having the vertices \(\displaystyle (\pm x,\,0)\) and \(\displaystyle (0,\,\p,1)\).

Calculate: \(\displaystyle \L\,\lim_{x\to0^+}\,\frac{\text{perimeter of R}}{\text{perimeter of Q}}\)

                  A(0,1)
                  *
                / | \
            Z /   |   \ W
 (-x/2,1/2) * - - + - - * (x/2,1/2)
          / :     |     : \
        /   :     |     :   \
  - C * - - + - - + - - - + - * B -
 (-x,0) \   :     |     :   / (x,0)
          \ :     |     : /
(-x/2,-1/2) * - - + - - * (x/2,-1/2)
            Y \   |   / X
                \ | /
                  *
                  D(0,-1)

Quadrilateral Q = ABCD.
Its side is: \(\displaystyle \,AB\:=\:\sqrt{(x-0)^2\,+\,(0,-1)^2}\:=\:\sqrt{x^2\,+\,1}\)
ABCD is a rhombus, so its perimeter is: \(\displaystyle \,P_{_Q}\:=\:4\sqrt{x^2\,+\,1}\)


W,X,Y,Z are the midpoints of the sides of Q.
Hence, rectangle R = WXYZ.
Its sides are: \(\displaystyle \,ZW\,=\,x\,\) and \(\displaystyle \,WX\,=\,1\)
Hence, its perimeter is: \(\displaystyle \,P_{_R} \;= \;2(x\,+\,1)\)

The ratio of the perimeters is: \(\displaystyle \L\,\frac{P_R}{P_Q}\;=\;\frac{2(x\,+\,1)}{4\sqrt{x^2\,+\,1}} \;=\;\frac{x + 1}{2\sqrt{x^2\,+\,1}}\)


Take the limit: \(\displaystyle \L\,\lim_{x\to0^+}\,\frac{x\,+\,1}{2\sqrt{x^2\,+\,1}}\;=\;\frac{0\,+\,1}{2\sqrt{0^2\,+\,1}}\;=\;\frac{1}{2}\)