exploring limits

Using Geometer’s Sketchpad to Support Mathematical Thinking

Shelly Berman p. 1 of 7 Jo Ann Fricker Exploring Limits.doc

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. 13 #60

Problem #2: Calculus: Graphical, Numerical, Algebraic By Finney, Thomas, Demana, Waits ©1995 Addison-Wesley Publishing Company, Inc. Page 116 #77



Problem #1:

C1

C2

Calculus with Early Transcendentalsby James Stewart©2003 Thomson Learning, Inc.Page 113 #60

What happens to R as C2 shrinks, that is, as r → 0+?

The figure shows a fixed circle C1

with equation (x – 1)2 + y2 = 1 and a shrinking circle C2 with radius rand center (0,0).

P is the point (0,r);Q is the upper point ofintersection of the two circles; R is the point of intersection of line PQ and the x–axis.

R

Q

(0,0) (1,0)

P

Animation #1:

c2

c1

Show Coordinate

Starting Position

Initial Condition

Shrink Radius

R

Q

(0,0) (1,0)

P



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+ .

The coordinates of

€

P are

€

0,r( ) . The point

€

Q is the point of intersection of the

two circles

€

x2 + y2 = r2 and

€

x −1( )2+ y2 =1.

Eliminating

€

y from these equations, we get

€

r2 − x2 =1− x −1( )2

r2 − x2 =1− x2 + 2x −1

r2 = 2x

∴ x =12

r2

Substituting back into the equation of the shrinking circle to find the y–

coordinate, we get

€

12 r2( )

2+ y2 = r2

y2 = r2 1− 14 r2( )

y = r 1− 14 r2

for the positive y–value.

So the coordinates for point

€

Q are

€

12 r2 ,r 1− 1

4 r2( ).

The equation of the line joining

€

P and

€

Q is thus

€

y − r =r 1− 1

4 r2 − r12 r2 − 0

⋅ x − 0( )



We set

€

y = 0 in order to find the x–intercept, and get

€

−r =r 1− 1

4 r2 −1( )12 r2 ⋅ x

Therefore,

€

x = −r ⋅12 r2

r 1− 14 r2 −1( )

x =− 1

2 r2 1− 14 r2 +1( )

1− 14 r2 −1

x = 2 1− 14 r2 +1( )

Now, we take the limit as

€

r → 0+ :

€

r→0+lim x =

r→0+lim2 1− 1

4 r2 +1( )

r→0+lim x =

r→0+lim2 1 +1( )

r→0+lim x = 4

So, the limiting position of

€

R is the point

€

4,0( ).

Calculus with Early Transcendentals Solution Manual By James Stewart ©2003 Thomson Learning, Inc.



Geometric Solution #1:

c2

c1

Show Coordinate

Starting Position

Initial Condition

Shrink Radius

T

S

R

Q

O (1,0)

P

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 .

Also

€

∠PSQ = ∠ORP since each angle is complementary to

€

∠SPQ.

Since

€

ΔQOS is isosceles, so is

€

ΔQTR, implying that

€

QT = TR. As the circle

€

C2 shrinks, the point

€

Q plainly approaches the origin, so the point

€

R must

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.



Problem #2:

Evaluate limP→Ob.

Let O denote the origin.

Let P(a,a2) be a point on theparabola y = x2 with a > 0.

f x( ) = x2

Let B(0,b) denote the y–intercept of theperpendicular bisector of line segment OP.

Calculus: Graphical, Numerical, Algebraicby Finney, Thomas, Demana, Waits©1995 Addison-Wesley Publishing Company, Inc.Page 116 #77

B(0,b)N

O

P

Animation #2:

a = 2.00f x( ) = x2

Show Intercept

Show Perpendicular Bisector

Show Slopes

Explore Limit

Starting Position

Coordinates of P

Animate Point

BN

P

O a



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+.

The coordinates of

€

O are

€

0,0( ) . The coordinates of

€

P are

€

a,a2( ). The point

€

N is the midpoint of

€

OP . Thus

€

N has the coordinates

€

0 + a2

,0 + a2

2

a2

,a2

2

The slope of

€

OP can be found

€

mOP

=a2 − 0a − 0

mOP

= a

Therefore, determining the equation of

€

BN requires the slope of the

perpendicular to

€

OP and its midpoint

€

N .

€

y −a2

2= −

1a

x −a2

y = −1a

x +12

+a2

2

Now, we take the limit as

€

a → 0+ of the y - intercept:

€

lima→0+

y = lima→0+

12

+12

a2

lima→0+

y =12

So, the limiting position of the

€

B is the point

€

0,12

.

exploring limits

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