Exploring Limits

Using Geometer’s Sketchpad to Support Mathematical Thinking
Exploring Limits
The concept of a limit is fundamental to Calculus. In fact, Calculus without limits is
like Romeo without Juliet. It is at the heart of so many Calculus concepts like the
derivative, the integral, etc. So what is a limit?
Maybe the best example to illustrate limits is through average and instantaneous
speeds: Let us assume you are traveling from point A to point B while passing through
point C. Then we know how to compute the average speed from A to B: it is simply the
ratio between the distance from A to B and the time it takes to travel from A to B.
Though we know how to compute the average speed this has no real physical
meaning.
Indeed, let us suppose that a policeman is standing at point C checking for speeders
going through C. Then the policeman does not care about the average speed. He only
cares about the speed that you see on the speedometer, the one that the car actually
has when crossing C. That one is real.
How do we compute this "instantaneous speed"? That's not easy at all! Naturally one
way to do this is to compute the average speed from C to points close to C. In this
case, the distance between these points and C is very small as well as the time taken
to travel from them to C. Then we look at the ratio: Do these average speeds over
small distances get close to a certain value? If so, that value should be called be the
instantaneous speed at C. In fact, this is exactly how the policeman's radar computes
the driver's speed!
http://www.sosmath.com/calculus/limcon/limcon01/limcon01.html
Calculus is built on the concept of limit. The rules for calculating limits are
straightforward, and most of the limits we need can be found by using one or
more strategies of direct substitution, graphing, calculator approximation, or
algebra. Here, we present Geometer’s Sketchpad animations created to explore
the algebraic and geometric aspects of two problems that nicely elucidate the
concept of limit.
Problem #1:
Calculus with Early Transcendentals
By James Stewart
©2003 Thomson Learning, Inc.
Page 113 #60
Problem #2:
Calculus: Graphical, Numerical, Algebraic
By Finney, Thomas, Demana, Waits
©1995 Addison-Wesley Publishing Company, Inc.
Page 116 #77
Shelly Berman
p. 1 of 7
Exploring Limits.doc
Jo Ann Fricker
Using Geometer’s Sketchpad to Support Mathematical Thinking
Problem #1:
P
Q
C2
The figure shows a fixed circle C1
with equation (x – 1)2 + y 2 = 1 and
a shrinking circle C2 with radius r
and center (0,0).
(1,0)
(0,0)
R
C1
Calculus with Early Transcendentals
by James Stewart
©2003 Thomson Learning, Inc.
Page 113 #60
P is the point (0,r);
Q is the upper point of
intersection of the two circles;
R is the point of intersection
of line PQ and the x–axis.
What happens to R as C2 shrinks, that is, as r → 0+?
Animation #1:
Starting Position
P
Initial Condition
c2
Q
(0,0)
(1,0)
R
c1
Shrink Radius
Shelly Berman
Show Coordinate
p. 2 of 7
Exploring Limits.doc
Jo Ann Fricker
Using Geometer’s Sketchpad to Support Mathematical Thinking
Algebraic Solution #1:
First we find the coordinates of P and Q as functions of r . Then we can find
the equation of the line determined by these two points, and thus find the x–
+
intercept (the point R ), and take the limit as r → 0 .
€
€
€
(
)
2
2
2
2
2
two circles€x + y = r and ( x −1)€ + y = 1.
The coordinates of P are 0,r . The point Q is the point of intersection of the
Eliminating y from these equations, we get
€
€
€
2
2
r − x = 1 − ( x −1)
€
r 2 − x 2 = 1 − x 2 + 2x −1
2
€
€
r 2 = 2x
1
∴ x = r2
2
Substituting back into the equation of the shrinking circle to find the y–
€
coordinate, we get
(
1
2
2
r 2 ) + y2 = r 2
y2 = r 2 (1 − 41 r 2 )
y = r 1 − 41 r 2
for the positive y–value.
€
So the coordinates for point Q are
(
1
2
)
r 2 ,r 1 − 41 r 2 .
The equation of the line joining P and Q is thus
€
r 1− 1 r2 €
− r
4

 ⋅ x − 0)
y−r =
 1 r2 − 0  (
 2 €
€
€
Shelly Berman
p. 3 of 7
Exploring Limits.doc
Jo Ann Fricker
Using Geometer’s Sketchpad to Support Mathematical Thinking
We set y = 0 in order to find the x–intercept, and get
−r =
(
)⋅x
r 1 − 41 r 2 −1
1
2
€
r
2
Therefore,
€
1
2
x = −r ⋅
x=
− 21
r2
(
r ( 1−
)
+ 1)
r 1 − 41 r 2 −1
2
1
4
r2
1 − 41 r 2 −1
(
)
x = 2 1 − 41 r 2 + 1
+
Now, we take the limit as r → 0 :
€
lim x = lim 2(
1 − 41 r 2 + 1
€ 2(
lim x = lim
1 +1
r →0 +
r →0
+
r →0 +
r →0
+
)
)
lim x = 4
r →0 +
(
)
So, the limiting position of R is the point 4,0 .
€
€
€
Calculus with Early Transcendentals
Solution Manual
By James Stewart
©2003 Thomson Learning, Inc.
Shelly Berman
p. 4 of 7
Exploring Limits.doc
Jo Ann Fricker
Using Geometer’s Sketchpad to Support Mathematical Thinking
Geometric Solution #1:
Starting Position
Initial Condition
P
c2
Q
O
T
(1,0)
R
c1
S
Show Coordinate
Shrink Radius
Now we add a few lines to the diagram, as shown. Note that ∠OQT = 90° and
∠PQS = 90 ° since each angle is inscribed in a semicircle. So ∠SQR = 90°
since it is supplementary to ∠PQS .
€
€
It follows that ∠OQS = ∠TQR since each angle is complementary
to ∠SQT .
€ since each angle is complementary to ∠SPQ .
Also ∠PSQ = ∠ORP
€
€ . As the circle
Since€ΔQOS is isosceles, so is ΔQTR , implying that QT = TR
€
€
€
so the point R must
C2 shrinks, the point Q plainly approaches the origin,
approach a point twice as far from the origin as T , that is, the point (4,0) .
€
€
€
€
€
€
€
Calculus with Early Transcendentals
Solution Manual
By James Stewart
©2003 Thomson Learning, Inc.
Shelly Berman
p. 5 of 7
Exploring Limits.doc
Jo Ann Fricker
Using Geometer’s Sketchpad to Support Mathematical Thinking
Problem #2:
f ( x ) = x2
Let P(a,a2) be a point on the
parabola y = x2 with a > 0.
P
B(0,b)
N
Let O denote the origin.
O
Calculus: Graphical, Numerical, Algebraic
by Finney, Thomas, Demana, Waits
©1995 Addison-Wesley Publishing Company, Inc.
Page 116 #77
Let B(0,b) denote the y–intercept of the
perpendicular bisector of line segment OP.
Evaluate limP→O b.
Animation #2:
P
f ( x ) = x2
Starting Position
a = 2.00
Animate Point
Show Slopes
B
N
Coordinates of P
Show Perpendicular Bisector
Explore Limit
Show Intercept
Shelly Berman
O
a
p. 6 of 7
Exploring Limits.doc
Jo Ann Fricker
Using Geometer’s Sketchpad to Support Mathematical Thinking
Algebraic Solution #2:
Since we are given the coordinates of O and P , we can find the coordinates of
N as a function of a. The perpendicular bisector of OP passes through N
and intersects the y–axis at point B . So we must find the equation for BN
+
€
€
and take the limit as a → 0 .
€
€
€
€2 ) . The point
The coordinates of O are€(0,0) . The coordinates of P are ( a,a
€
N is the midpoint of OP . Thus N has the coordinates
€
 0 + a 0 + a2 
 € , € 
2 
 2
€
€
 a a2 
 , 
2 2 
€
€
€
The slope of OP can be found
€
mOP
€ m
OP
a2 − 0
=
a−0
=a
Therefore, determining the equation of BN requires the slope of the
€
perpendicular to OP and its midpoint N .
a2
1
a
y−
= − x −  €
2
a
2
€
 1 a2  €
1
y=− x+ +

a
2 2 
+
Now, we take the limit as a → 0 of the y - intercept:
€
1 1 
lim y = lim + a2 
a→0
a→0  2
2 
€
1
lim y =
a→0
2
+
+
+

1
2
So, the limiting position of the B is the point  0,  .

€
Shelly Berman
€
p. 7 of 7
Exploring Limits.doc
€
Jo Ann Fricker