LIMITSPatrick Dale McCray, Ph.D.1 September, 20091 Transforming lives. Inventing the future. www.iit.eduLIMITSWhy limits?What are they?How can they be found?Limits are certain kinds of numbers associated with functions.Unsure of what functions are? Then find out now, and come back to this workshop later. A goodplace to start is Stewart’s Calculus, 6th edition. Take the diagnostic test on functions. Then review theideas about functions in sections 1.1, 1.2, and 1.3.Limits are used for figuring out many things, such as slopes and speeds. What is the slope of the tangent to a curve at a given point? How fast is something moving at a specific time?First, we will look at how limits help us understand the tangent and velocity problems. Then we willexamine intuitively the idea of the limit of a function at a point. Finally, we conclude with basic ways forfinding limits of simple functions.2 Transforming lives. Inventing the future. www.iit.eduTHE TANGENT PROBLEMWhen a ball is resting on the floor, the flat surface of the floor is tangent to the ball at the point ofcontact. If you put a piece of wire on the floor shaped to resemble a circle, say, and slide a ruler up to thewire in such a way as to touch the wire in only one point, you’d have another example of a tangent.Consider the curve C given by the formulaC W y D f.x/ Dx34:We can use the point-slope formula` W y  b D m .x  a/:for a line ` through a point .a; b/ with slope m to find the tangent lines for the curve C.To be specific, let us find the equation of the line tangent to our cubic curve at the point where x D 2.When x D 2 the value of the function is y D f.2/ D 23=4 D 2. Therefore the point where the line ` istangent to our curve is P .2; 2/, so the only thing we have to do is find the slope m.3 Transforming lives. Inventing the future. www.iit.eduRight now we do not have a way of directly determining the slope m. But what we can do is find theslope of a straight line between any two points. Recall that if .x1; y1/ and .x2; y2/ are the coordinates oftwo points, the slope of the line connecting those two points is given by the change in y divided by thechange in x. namelym Dy2 y1x2 x1:So, if we let P be the point P .2; 2/ and if we let Q be a nearby point on the curve given by Q.x; y/ D.x; x3=4/, the closer x is to 2, that is, the closer the point Q is to P , the closer the secant through P andQ gets to the tangent to the curve at P.–50510151234x4 Transforming lives. Inventing the future. www.iit.eduNow the slope mPQof the secant between points P and Q is given bymPQDy  2x  2Dx3=4  2x  2:To find out what happens to the slope of the secants when x increases from 1 to 2 we can compute a fewvalues. From the graph we would expect these slopes to increase as x approaches 2. We can performsimilar calculations on the right, as x decreases from, say, 3 to 2. These slopes should be decreasing.x mPQ1 1.751.5 2.31251.9 2.85251.99 2.9850251.999 2.9985x mPQ3 4.752.5 3.81252.1 3.15252.01 3.0150252.001 3.0015From the tables we see that the limit from the left as x goes to 2 appears to be 3. We also see that the limitfrom the right as x goes to 2 also appears to be 3. Since these two limits are equal, we guess that 3 is thevalue of the slope of the tangent line at x D 2. Therefore the equation of the tangent line in point-slopeform isy  2 D 3 .x  2/:5 Transforming lives. Inventing the future. www.iit.eduCHALLENGE 1Now it is your turn!Find the tangent to the curve y D x2at x D 3 using what you’ve just learned about limits.0102030123456x6 Transforming lives. Inventing the future. www.iit.eduTHE VELOCITY PROBLEMIf a car is driven around a race track with a constant velocity, then it is a simple matter to find out thedistance traveled, if you know how long the car has been going, by using the formuladistance D velocity timeor, symbolically, ass D v t :Inverting the problem in this situation we can infer how fast the car was going ( v ), given how long( t ) the car took to travel the distance ( s ), by solving the equation for the velocity v, as a function of thedistance s and time t asv Dst:But what if instead of moving at a constant speed the car is being driven across town in moderate traffic?There will be periods of time when the car is not moving. Occasionally the car will be moving at the speedlimit. But more likely the car will be moving less than the speed limit most of the time. And that meansthat the car will take longer to travel a specific distance than if it had been able to travel at its maximumspeed the whole time, that is, at the speed limit. So, the velocity is varying, depending upon local trafficconditions at any given point in time.7 Transforming lives. Inventing the future. www.iit.eduWhen we look at the speedometer we see how fast the car is going at that point in time. Well supposefor a few seconds the velocity is not changing. If we divide the distance traveled in those few seconds bythe time elapsed, we would get the average velocity for that time period. Let us denote the two times as t1and t2with t1< t2. Denote by s1the distance reading on the odometer at time t1. Denote by s2the readingon the odometer at time t2. Thenaverage velocityŒt1;t2Dchange in positionchange in timeDs2 s1t2 t1:We’d expect the result to agree with the velocity shown on the speedometer. So, to find out howfast the car is moving more accurately, we could compute the average velocities over shorter and shorterintervals of time and see what number they appear to be approximating (approaching). As t1approachest2, that limiting velocity would be the instantaneous velocity of the car at that point in time, namely t2.instantanous velocity at t2Dlimitt1! t2s2 s1t2 t1:To make things more concrete, we next study the situation of a body falling near the surface of theearth. For objects in free fall, the height of the object, or the distance traveled, is a known function of thetime elapsed. This will allow us do some simple arithmetic calculations.8 Transforming lives. Inventing the future. www.iit.eduThe formula for the distance a falling object travels in t seconds is given by s.t / D12gt2: Theinternational standard value of the acceleration g due to gravity near the earth’s surface is 980.665 cm/sec2.Converting this value to feet per second squared gives a value of approximately32:1740485564304132480 ::: ft/sec2for the acceleration g due to gravity.. That’s pretty messy, isn’t it? For