2.1 The Tangent and Velocity Problems

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. The Tangent and Velocit Problems Suppose we want to find the slope of the line between (, ) and (-, -). We could do this easil with the slope formula: below. m = = =. Now consider the function =. Suppose we want to find the slope of the secant line shown 5 3 5 3 We know two points on the line so we can find the slope as we before. What about the slope of the tangent line at (, 0)? We don t know two points so we can t calculate it. What we can do is approimate the slope b taking a second point which is close to (, 0) and calculating the slope of the secant line. The closer we get to (, 0) the closer our approimation will be to the true slope. For eample we could tr find the slope between the points (, ) and (,0). The slope of the secant line between those two points would be What if we tr some values of close to? m = 0..5.9..5 3 m =

Eample... The point P(0.5, ) lies on the curve = /. If Q is the point (,/), use our calculator to find the slop of the secant line PQ for the following values of. a).8 b).6 c).5 d).9 e).5 Guess the value of the slope of the tangent line at P(0.5,). Average Velocit Average Velocit = change in distance change in time Distance is usuall given as a function of time s = s(t). This function tells us that if we know how long the object has been moving, t, we know where it is in space s(t). Eample... Suppose that the position of an object moving in a straight line is given b the equation s = t 3 /6 where t is in seconds, s in meters. Find the average velocit over the given time periods then find the instantaneous velocit when t =.. [,3]. [,] 3. [,.5]. [,.]

. The Limit of a Function The Question: What happens to f() as approaches a? The Mathematical notation: lim f() = L In words: The limit of f() as approaches a, equals L Consider the following functions 6 6 3 3 f() = + f() = g() = g() is undefined As gets close to, As gets close to, = f() gets close to. = g() gets close to. Translation: lim f() = lim g() = So when we write lim f() = L what we mean in English is As gets CLOSE to a then f() gets CLOSE to L. In Words: So when we write lim f() = L what we sa out loud is The limit as gets approaches a of f() is L. The ke word here is CLOSE. 3

Consider the following graph: 3 3 5 6 7 Evaluate the following: (a) lim f() = (b) lim f() = (c) lim f() = 3 (d) f() = (e) f() = (f) f(3) = One Sided Limits The limit as approaches a from the left: lim f() = L The limit as approaches a from the right: lim +f() = L The limit eists if and onl if the limit from the left is equal to the limit from the right. Infinite Limits lim f() = L lim = lim f() +f() = L For eample f() = tan or f() = (+) lim f() = ±

Eample... Sketch the graph of the following function and use it to determine the values of a for which lim f() eists: if < f() = if < ( ) if 3 3 Eample... Sketch the graph of a function f that satisfies these conditions: lim =, 3 +f() lim =, 3 f() lim f() =, f(3) = 3, f() =. 5

Eample..3. Determine the infinite limits: 6 a) lim 5 + 5 6 b) lim 5 5 c) lim 0 (+) Eample... Determine the limits from the graph of f(). 5 3 5 3 5 a) lim 3 +f() = b) lim 3 f() = c) lim +f() = d) lim f() = 6

.3 Calculating Limits Using the Limit Laws Recall: Notation lim f() = L means as gets close to a, f() gets close to L. Limits have certain properties which make them eas to work with. Properties: Suppose then: lim f() = L and lim g() = K, ) lim b f() = b L ) lim [f()±g()] = L±K 3) lim [f() g()] = L K [ ] f() ) lim = L g() K when K 0 5) lim [f()] n = L n We can use these properties to help us solve limits analticall (using algebra). Eample.3.. lim +7+ EVERY time ou get to a limit problem the first thing ou alwas want to do is to plug in the value that ou are given. STEP : Alwas plug in the value. f() = + + = 0 If ou get a number, as in this case, then the problem is done. Here the answer is 0. As gets close to, +7+ gets close to 0 Eample.3.. lim + STEP : Plug in for : ( ) ( ) ( )+ = 0 0 Here we didn t get a number but rather an undefined epression of the form 0. This means that we can 0 do some algebra and cancel some common factors. 7

Step : Do some algebra and cancel. + Eample.3.3. lim 0 +0 STEP : Plug in for : 0 = 0 0 Here we didn t get a number but rather an undefined epression of the form 0. This means that we can 0 do some algebra and cancel some common factors. Step : Do some algebra and cancel. In this case we can t factor so we will multipl b the conjugate of the numerator to get rid of the square roots. This is known as rationalizing the numerator. 8

Eample.3.. lim 3 9 t 3 t Eample.3.5. lim 0 + Eample.3.6. lim f() if f() = 3+9 if < if > 9

Eample.3.7. lim h 0 (+h) h STEP : Plug in for h = 0: Here we didn t get a number but rather an undefined epression of the form 0. This means that we 0 can do some algebra and cancel some common factors. Step : Do some algebra and cancel. In this case we will multipl out the numerator. Eample.3.8. Find lim h 0 f(+h) f() h where f() =. f() f(+h) {}}{{}}{ f(+h) f() (+h) (+h) ( ) lim = lim h 0 h h 0 h 0

The Squeeze Theorem If f() g() h() when is near a (ecept possibl a) and lim f() = lim h() = L then ( ) Eample.3.9. Evaluate lim sin 0 lim g() = L f() = 0.5 0.0.0 0.5 0.05 0.05 0.0 0.5 0.5.0 h() = 0.05 f() = 0. 0. 0. 0. 0.3 0.05 h() =

. The Precise Definition of a Limit Recall: A. Informal Definition of Limit Let f() be defined on an open interval about a ecept possibl at a itself. If f() gets arbitraril close to L (as close to L as we like) for all sufficientl close to a, we sa that f approaches the limit L as approaches a and we write: lim f() = L ( the limit of f(), as approaches a, equals L ). Notes: () a means that ou approach = a from both sides of a. () f(a) does not have to be defined. B. Solving an inequalit. Fill in the blanks: To sa that < means that is less than units from.. Solve < Definition.. The ǫ δ Definition of Limit Let f() be defined on an open interval about a ecept possibl at a itself. Then we sa that f approaches the limit L as approaches a and we write: lim f() = L if for all ǫ > 0 there eists a δ > 0 such that f() L < ǫ whenever 0 < a < δ. OR if 0 < a < δ then f() L < ǫ.

Eample... Consider the function f() = +5. How close to a =, must we hold to be sure that f() lies within.5 units of f(a) = 3? In terms of the definition f() = +5, a =, L = 3 and ǫ =.5 the distance between and OR the distance between and is the desired value of δ. 6 L+ǫ L L ǫ 5 3 3 We can also use algebra. We want to know when is f() L <.5? We solve the inequalit: f() <.5 (+5) <.5 Eample... Use the graph of f() = 9, a = 0, L = 3 and ǫ = to determine the value of δ using the definition of lim 9 = 3. To do this we will find a number δ such that if 0 < 0 < δ 0 then 9 <. 5 L+ǫ L 3 L ǫ a δ 0 a+δ 0 3

If the value of ǫ is not specified then ou solve for δ in terms of ǫ. This usuall involves using algebra on the inequalit f() L < ǫ until ou can get it to look like a < something. At that point something = δ. Eample..3. Prove that lim 3+5 = 8. Given an ǫ > 0 we need to find δ > 0 such that 0 < a < δ = f() L < ǫ. 0 < < δ = (3+5) 8 < ǫ = 3 < ǫ = 3 < ǫ = < ǫ 3 This suggests that we choose δ = ǫ 3. Let s show that δ = ǫ 3 works in the definition. If 0 < < δ = ǫ 3 then ( ǫ (3+5) 8 = 3 = 3 < 3δ = 3 = ǫ. 3) Eample... The interior of a tpical -L measuring cup is a right circular clinder of radius 6cm. How closel must we measure the height, h, in order to measure out L (000 cm 3 ) with an error of no more than % (i.e. 0 cm 3 )? (Use: V = πr h)

Eample..5. Use the Graph of f() = to find a number δ such that < 0.5 whenever < δ. 6 L+ǫ L L ǫ 5 3.0 0.5 0.5.0.5.0.5 3.0 5