Exploring Limits

Using Geometer’s Sketchpad to Support Mathematical ThinkingExploring LimitsThe concept of a limit is fundamental to Calculus. In fact, Calculus without limits islike Romeo without Juliet. It is at the heart of so many Calculus concepts like thederivative, the integral, etc. So what is a limit?Maybe the best example to illustrate limits is through average and instantaneousspeeds: Let us assume you are traveling from point A to point B while passing throughpoint C. Then we know how to compute the average speed from A to B: it is simply theratio 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 physicalmeaning.Indeed, let us suppose that a policeman is standing at point C checking for speedersgoing through C. Then the policeman does not care about the average speed. He onlycares about the speed that you see on the speedometer, the one that the car actuallyhas when crossing C. That one is real.How do we compute this "instantaneous speed"? That's not easy at all! Naturally oneway to do this is to compute the average speed from C to points close to C. In thiscase, the distance between these points and C is very small as well as the time takento travel from them to C. Then we look at the ratio: Do these average speeds oversmall distances get close to a certain value? If so, that value should be called be theinstantaneous speed at C. In fact, this is exactly how the policeman's radar computesthe driver's n01/limcon01.htmlCalculus is built on the concept of limit. The rules for calculating limits arestraightforward, and most of the limits we need can be found by using one ormore strategies of direct substitution, graphing, calculator approximation, oralgebra. Here, we present Geometer’s Sketchpad animations created to explorethe algebraic and geometric aspects of two problems that nicely elucidate theconcept of limit.Problem #1:Calculus with Early TranscendentalsBy James Stewart 2003 Thomson Learning, Inc.Page 113 #60Problem #2:Calculus: Graphical, Numerical, AlgebraicBy Finney, Thomas, Demana, Waits 1995 Addison-Wesley Publishing Company, Inc.Page 116 #77Shelly Bermanp. 1 of 7Exploring Limits.docJo Ann Fricker

Using Geometer’s Sketchpad to Support Mathematical ThinkingProblem #1:PQC2The figure shows a fixed circle C1with equation (x – 1)2 y 2 1 anda shrinking circle C2 with radius rand center (0,0).(1,0)(0,0)RC1Calculus with Early Transcendentalsby James Stewart 2003 Thomson Learning, Inc.Page 113 #60P is the point (0,r);Q is the upper point ofintersection of the two circles;R is the point of intersectionof line PQ and the x–axis.What happens to R as C2 shrinks, that is, as r 0 ?Animation #1:Starting PositionPInitial Conditionc2Q(0,0)(1,0)Rc1Shrink RadiusShelly BermanShow Coordinatep. 2 of 7Exploring Limits.docJo Ann Fricker

Using Geometer’s Sketchpad to Support Mathematical ThinkingAlgebraic Solution #1:First we find the coordinates of P and Q as functions of r . Then we can findthe 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 . ()22222two 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 theEliminating y from these equations, we get 22r x 1 ( x 1) r 2 x 2 1 x 2 2x 12 r 2 2x1 x r22Substituting back into the equation of the shrinking circle to find the y– coordinate, we get(122r 2 ) y2 r 2y2 r 2 (1 41 r 2 )y r 1 41 r 2for the positive y–value. So the coordinates for point Q are(12)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 Bermanp. 3 of 7Exploring Limits.docJo Ann Fricker

Using Geometer’s Sketchpad to Support Mathematical ThinkingWe set y 0 in order to find the x–intercept, and get r () xr 1 41 r 2 112 r2Therefore, 12x r x 21r2(r ( 1 ) 1)r 1 41 r 2 1214r21 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 lim1 1r 0 r 0 r 0 r 0 ))lim x 4r 0 ()So, the limiting position of R is the point 4,0 . Calculus with Early TranscendentalsSolution ManualBy James Stewart 2003 Thomson Learning, Inc.Shelly Bermanp. 4 of 7Exploring Limits.docJo Ann Fricker

Using Geometer’s Sketchpad to Support Mathematical ThinkingGeometric Solution #1:Starting PositionInitial ConditionPc2QOT(1,0)Rc1SShow CoordinateShrink RadiusNow 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 complementaryto SQT . since each angle is complementary to SPQ .Also PSQ ORP . As the circleSince ΔQOS is isosceles, so is ΔQTR , implying that QT TR so the point R mustC2 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 TranscendentalsSolution ManualBy James Stewart 2003 Thomson Learning, Inc.Shelly Bermanp. 5 of 7Exploring Limits.docJo Ann Fricker

Using Geometer’s Sketchpad to Support Mathematical ThinkingProblem #2:f ( x ) x2Let P(a,a2) be a point on theparabola y x2 with a 0.PB(0,b)NLet O denote the origin.OCalculus: Graphical, Numerical, Algebraicby Finney, Thomas, Demana, Waits 1995 Addison-Wesley Publishing Company, Inc.Page 116 #77Let B(0,b) denote the y–intercept of theperpendicular bisector of line segment OP.Evaluate limP O b.Animation #2:Pf ( x ) x2Starting Positiona 2.00Animate PointShow SlopesBNCoordinates of PShow Perpendicular BisectorExplore LimitShow InterceptShelly BermanOap. 6 of 7Exploring Limits.docJo Ann Fricker

Using Geometer’s Sketchpad to Support Mathematical ThinkingAlgebraic Solution #2:Since we are given the coordinates of O and P , we can find the coordinates ofN as a function of a. The perpendicular bisector of OP passes through Nand intersects the y–axis at point B . So we must find the equation for BN and take the limit as a 0 . 2 ) . The pointThe 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 mOPa2 0 a 0 aTherefore, determining the equation of BN requires the slope of the perpendicular to OP and its midpoint N .a21 a y x 2a 2 1 a2 1y x a 2 2 Now, we take the limit as a 0 of the y - intercept: 1 1 lim y lim a2 a 0a 0 22 1lim y a 02 1 2 So, the limiting position of the B is the point 0, . Shelly Berman p. 7 of 7Exploring Limits.doc Jo Ann Fricker

By Finney, Thomas, Demana, Waits 1995 Addison-Wesley Publishing Company, Inc. Page 116 #77. Using Geometer's Sketchpad to Support Mathematical Thinking Shelly Berman p. 2 of 7 Jo Ann Fricker Exploring Limits.doc Problem #1: C1 C2 Calculus with Early Transcendentals by James Stewart